2E NBDEI NNGD I N G 290 CHAPTER 6 M (kN"m) BENDING The simply supported beam in Fig. 6–26a has the cross-sectional area 22.5 shown in Fig. 6–26b. Determine the absolute maximum bending stress The simply supported beam in Fig. 6–26a has the cross-sectional area in the beam and draw the stress distribution over the cross section at EXAMPLE 6.12 2 shown 2 9 0 in Fig.C6–26b. H A P T E RDetermine 6 B E N D I Nthe G absolute maximum bending stress this location. in the beam and draw the stress distribution over the cross section at The simply supported beam in Fig. 6–26a has the cross-sectional area 6m this location. TheThe simply supported beam inmFig. 6–26a has the cross-sectional areaarea 5beam kN/ Thehas simply supported in Fig. 6–26a has the cross-sectional area simply supported in Fig. 6–26a the cross-sectional shown in Fig. 6–26b. Determine the absolute maximum bending stress beam (a) EXAMPLE 6.12 Mthe (kN"m) shown in Fig. 6–26b. Determine the absolute maximum bending stress shown in Fig. 6–26b. the maruz absolute maximum bending stress shown in Fig. 6–26b. Determine absolute maximum bending stress ! and Şekilde enkesiti üniform yayılı yüke basit mesnetli kirişte in the beam stress distribution over the verilen cross section atDetermine 5draw kN/ mthe in the beam and draw the stress distribution over the cross section at in the beam and draw the stress distribution over the cross section at in the beam and draw the stress distribution over the cross section at SOLUTION this location. M (kN"m) oluşacak mutlak değerce maksimum eğilme gerilmesini belirleyerek enkesit üzerindeki 22.5 thisthis location. this location. location. The simply supported beam in Fig. 6–26a has the cross-sectional area Momen Maximum Internal dağılımını gösteriniz. 290 C H A P T E22.5 R 6 BENDING 20 mm Determine the absolute maximum 5 kN/m shown in Fig. 6–26b. bending stress the beam, M = 22.5 kN # m, o 5 kN/ m kN/ m 5 kN/m M 5(kN"m) in the beamMand draw the stress distribution x (m) over the cross section at B M (kN"m) M (kN"m) (kN"m) 3 6150 mm Section Property. By reaso C 290 C H A P T E R 6 B E N D I N G this location. 6m 6 x (m) N A through the centroid C at the EXAMPLE 6.12 22.5 22.5 22.5 3 620 mm (a) 22.5(c) area is subdivided into the t 5 kN/ m 6m 150 mm M (kN"m) inertia of each part is calcu SOLUTION (a) EXAMPLE 6.12 (c) 20 mm The simply supported beam in Fig. 6–26a has the cross-sectional area theorem. (See E x (m) x (m) parallel-axis xinternal (m) x (m) Maximum Internal The maximum moment in D 3 6 3 Moment. 6 SOLUTION shown in Fig. 6–26b. Determine the absolute maximum bending stress 3# 6 22.5 3 work in6 meters, we have m 6m the beam, occurs at the center. M = 22.5 kN m, 250 mm 6m in the beam andinternal draw the stress section at area The simply supported beamdistribution in Fig. 6–26aover has the the cross cross-sectional Maximum Internal Moment. The maximum moment (c) a) (a) B (c) (c) I = ©1I + Ad22 (a) 150 mm (c) # this location. shown in Fig. 6–26b. Determine absolute maximum bending stress the center. 22.5 kN Property. m, occurs atBy reasons of symmetry, the neutral passes (b)the axis C the beam, M = Section x (m) SOLUTION in and draw stressFig. distribution over the cross section SOLUTION N 150 mm SOLUTION Çözüm: A through the centroid CSOLUTION at the the beam midheight of thethe beam, 6–26b. 3 The 6 1 at 20 mm 6m 5 kN/ m the neutral axis passes Section Property. Bysubdivided reasons of into symmetry, = 2c 10.25 m210.020 this area is thelocation. three parts shown, moment of Maximum Moment. TheThe maximum internal moment inand Maximum Internal Moment. The maximum internal moment inmoment Moment. maximum internal in the 12 in 150 mm Internal M (kN"m) Maximum Internal Moment. The maximum internal moment A Maximum 12.7 MPa through theInternal centroid C at the midheight of the beam, Fig. 6–26b. The (a) (c) #kN # # inertia each is calculated about the neutral axis using the the the beam, occurs atpart the center. M =M 22.5 kN m,of the beam, occurs at the center. M = 22.5 kN m, beam, occurs at the center. = 22.5 m, # 20 mm the shown, beam, at the center. M kN m, occurs 1 area is subdivided into the theorem. three parts and the moment of 5 kN/ m = 22.5 150 mm parallel-axis (See Eq. A–5 of Appendix A.) Choosing to + c 10.020 m210.300 SOLUTION D 22.5 MB 0 mm (Maksimum moment) 11.2 MPa 150 mm B (kN"m) inertia of each part is calculated about the neutral axis using the 12 150 mmof we Section Property. By reasons symmetry, the the neutral axisaxis passes Section Property. By reasons of have symmetry, neutral passes Section Property. By reasons symmetry, the neutral axis passes work inof meters, Property. ByMoment. reasons the neutral axis passes in C 250 mm parallel-axis theorem. Maximum Internal The maximum internal moment (See Eq. A–5ofSection ofthe Appendix A.) Choosing to of symmetry, A 6A through the centroid C at the midheight beam, Fig. 6–26b. The AD through through the centroid C at the midheight of the beam, Fig. 6–26b. The the centroid C at the midheight of the beam, Fig. 6–26b. The N A 2 through the centroid C at the = 301.3110 midheight of the beam, Fig. 6–26b. The -62 m4 # 20 mm the beam, occurs at the center. M = 22.5 kN m, I = ©1I + Ad 2 work meters, we into have 22.5 20 mm the M ! 22.5 kN"m is subdivided into the the three parts shown, and the moment of area isinsubdivided three parts shown, and the moment of area(b)isarea subdivided into three parts shown, and the moment of 0 mm x (m) 150 mm D area is subdivided into the three parts shown, and the moment 150150 mmmm B 22.5 Mc of 3 6 of is 2calculated the neutral axispart using the of©1I each is calculated the neutral axis the 1 about I part =each + part Ad inertiainertia of inertia each ispart calculated theabout neutral axis using the using Section Property. By reasons of symmetry, the neutral axis passes 3 each 2the neutral s = ; smax = inertia of is calculated about axis using the C= 2about 62m max c 10.25 m210.020 m2 + 10.25 m210.020 m210.160 m2 d 20 mm I The theorem. (See Eq. A–5 of through Appendix A.) Choosing to midheight parallel-axis (SeeA–5 Eq.of A–5 ofAAppendix A.) Choosing 6 parallel-axis N theorem. theorem. (See Eq. Appendix A.) Choosing to (See thetheorem. centroid Ctoat Eq. the of the beam, Fig. 6–26b. D D parallel-axis (m) parallel-axis A–5 (a) D 12 12.7 MPa (c) of Appendix A.) Choosingx to 201 mmhave 3 2 work in meters, we have work in meters, we 3 6 = 2 c 10.25 m210.020 m2 + 10.25 m210.020 m210.160 m2 d area is subdivided into the three parts shown, and the moment of work in meters, we have work in meters, we have A three-dimensional view 150 mm 1 6 m mm + c 12 250 10.020SOLUTION m210.300 m23ofd each part .7 MPa 2 2 inertia is calculated about the neutral axis using the 12.7 MPa Fig. 6–26d. Notice how the I = ©1I + Ad 2 2 11.2 MPa I = ©1I + Ad 2 2 12 mmAd 2 (a) (c) I = ©1I20 + I = ©1I + theorem. Ad 2 1 parallel-axis (See Eq. A–5 of Appendix A.) Choosing to 3 Maximum Internal Moment. The maximum internal moment in section develops a force th (b) D +1 c 1 10.020 m210.300 m2 -6 d 4 11.2 MPa 20 mm = 301.3110 2 mbeam, # 12 10.25 work inm210.160 meters, we have 2 SOLUTION 3m23 +the 2m2 occurs at the center. M = 22.5 kN m, 1= 2 = 1 neutral axis that has the same 2 c m210.020 10.25 m210.020 m210.160 d 250 mm c 10.25 m210.020 m2 + 10.25 m210.020 m2 d 2 ! 22.5 m210.020 kN"m = D 2c M 10.25 m23 4+ 10.25 m210.020 m210.160 d m210.020 = 2c 3 m2 10.25 m23 + 10.25 m210.020 m210.160 m22 d -6 B12 7 MPa 22.5(10 ) Internal N # +m10.170 yB = moment 150 mm, and Maximum The maximum internal in so Pa (d) Mc 12 =12301.3110 2 mmm I = 12 ©1I Ad2Moment. 2 m2 150 12.7 MPa Property. By reasons of symmetry, the neutral axis passes Ans. ; sSection = 12.7 MPa Cs(b) # max = max = 20 mm 1 M ! 22.5 kN"m 1 -6 4 the beam, occurs at the center. M = 22.5 kN m, 3 D 3m2 d3 I22.5(10 301.3110 m the midheight 1 1 C2 at 6 1 N+ c + c10.020 10.020 m210.300 m210.300 m2 d A )through N # m10.170 22. the centroid the beam, Fig. 6–26b. My TheB 2 3 Mc Fig.3 6–26 += c m2 10.020 m210.300 m2 d 3 +of10.25 11.2 11.2 MPaMPa + c s10.020 m210.300 d=150 12 ; B sm2 2012 mm Ans. = = 12.7 MPa 2 c 10.25 m210.020 m2 m210.020 m2; d sB = sm210.160 - of mm 11.2 MPa max max B a B = axis 12 -6 4 12 area is subdivided into the three parts shown, and the moment A three-dimensional view of the stress distribution is shown in Section Property. By reasons of symmetry, the neutral passes I 12 150 mm C12.7 I 4 MPa 301.3110 2 m -6 -642 m =N-6301.3110 6 2.7 MPa = 301.3110 m 6–26d. Noticeinertia of =stress each part is2 calculated about the cross neutral axis Fig. using the The A -6 4CBatand howthrough the at points D on the the centroid the midheight of the beam, 6–26b. 4 2Fig. 1 301.3110 m 20 mm = 301.3110 220mmm ! 22.5 kN"m A three-dimensional view3 of 3 the distribution is m210.300 shown A–5 inm23ofd Appendix A.) Choosing to M 22.5 kN"m D !M ccontributes 10.020 theorem. (See # stress section am10.170 that moment aboutshown, the and the moment of aream2 is +subdivided intoaEq. the three parts Mdevelops !11.2 22.5 kN"m 22.5(10 N force m10.170 m2 kN"m )MPa N )# parallel-axis mm D22.5(10 3 Mc Mc Notice B6–26d. 12 # m10.170 3 150 Fig. how the stress at points B and D on the cross # 22.5(10 ) Nat Ans. s = ; s = = 12.7 MPa Mc 22.5(10 ) N m10.170 m2 work in meters, we have Ans. sMc = ; s = = 12.7 MPa neutral axis that has the same direction as M. Specifically, point B, m2 max max inertia of each part is calculated about the neutral axis using the max max -64 4 -6 s 250=mm I Ans. = ; s = = 12.7 MPa 20 ;mm I sdevelops 301.3110 2 m a force that contributes a moment about the Ans. smax = section = 12.7 MPa max max -6 4 301.3110 2 m -6 4 max -6 4 = 301.3110 2 m and so y = 150 mm, I (d) parallel-axis theorem. (See Eq. A–5 of Appendix A.) Choosing to 301.3110 2 m Ineutral axis that Bhas301.3110 2direction mkN"m I = D + Ad22 the same as ©1I M. Specifically, at point in B, M ! 22.5 A three-dimensional of the stress distribution is shown D view 3 (b) A three-dimensional view of the stress distribution is shown in work in meters, we have # 3 m10.170 m2 Mc 250 mm and so yB =6–26d. 150 mm, A 22.5(10 three-dimensional view of the) N stress distribution is shown in ) N;D# m10.150 m222.5(10 Fig. 6–26 Fig. My A three-dimensional view of the stress distribution shown insthe B stress Notice at =points B on 1=isand sBmaxand =cross = 12.7 2MPa Ans. Fig. 6–26d. Notice how stress points D on the cross max 3 2 sB how =the - the ; at s = 11.2 MPa -6 4 12.7 MPa B Fig. = 2c 10.25 m210.020 m2 + 10.25 m210.020 m210.160 m2the d cross 6–26d. Notice how the stress at points B and D on I I = ©1I + Ad 2 -6 4 301.3110 2 m Fig. 6–26d. Notice how the at points on the cross 3 B and I22.5(10 section develops astress force that contributes a 301.3110 moment 2about m the the section develops a (b) force that contributes aDmoment about 12 ) N # m10.150 m2 6 My B 12.7 MPa section develops a force that contributes a moment about the sectionneutral develops athat force that aas moment about s the sB = -axis ; has =contributes 11.2 MPa neutral that same direction as M. three-dimensional Specifically, at point B, of the stress distribution is shown A axis M. point B,view 1 = at-the -6Specifically, 3 2 B, in 1=4axis I has theBsame direction 301.3110 2 m neutral that has the same direction as M. Specifically, at point 3 2c 10.25 m210.020 m2 + 10.25 m210.020 m210.160 m2 12.7 neutralyBaxis hasMPa the as M. Specifically, at12 point B, howm2thed stress at points B and D on the dcross and yBthat = 150 mm, + c 6–26d. 10.020 m210.300 Fig. Notice andsame so sodirection = 150 mm, 11.2MPa MPa B 12 and so y = 150 mm, 12.7 (d) B yB = 150 mm, and so section develops a force that contributes a moment about the 3 3 22.5(10 )# m10.150 N # m10.150 m2 1 -6 6 My 22.5(10 ) N m2 4 B 33 My neutral that has the22.5(10 same as M.m2 Specifically, at point B, = 301.3110 2 -m11.2 + caxis 10.020 m210.300 m2direction d) N # m10.150 s s = - B ; ; Fig. MPa 6–26 B-= 3)-N # sB22.5(10 sBMy =B =M =12 11.2 MPa B-= -64 4 My 11.2 MPa m10.150 m2 B -6 I ! 22.5 kN"m 301.3110 2B=m B s s ; = = -11.2 MPa and so y = 150 mm, I (d) D 301.3110 2 m 3 B B -6 m2 sB = sB = ; =Mc -11.2 ) 301.3110 N # m10.170 I MPa -6 22.5(10 2 m4 4 4 I 301.3110-62 ms = s =2 m = 12.7 MPa Ans. =; 301.3110 3 # 22.5(10-6 )2 N Fig. 6–26 M ! 22.5 kN"mmax I MyB max 301.3110 m4 m10.150 m2 D 3 sB = 22.5(10 sB = ; = - 11.2 MPa # ) N m10.170 -6 McI 2 m4m2 =is12.7 A three-dimensional view= of the 301.3110 stress -6 distribution shown smax = ; smax MPain Ans. 4 I 2 mand 12.7 MPa Fig. 6–26d. Notice how the stress 301.3110 at points B D on the cross 12.7 MPa (d) Fig. 6–26 (d) Fig. 6–26 sectionAdevelops a force that a moment about the in three-dimensional viewcontributes of the stress distribution is shown neutral that Notice has the how samethe direction Specifically, at on point Fig.axis 6–26d. stress as at M. points B and D theB,cross and so a force that contributes a moment about the yB =section 150 mm,develops neutral axis that has the same3 direction as M. Specifically, at point B, 22.5(10 ) N # m10.150 m2 MyB and so y = 150 mm, sB = sB = B; = - 11.2 MPa I 301.3110-62 m4 3 # 22.5(10 ) N m10.150 m2 MyB sB = sB = ; = - 11.2 MPa I 301.3110-62 m4 PAGE !82 6.4 THE FLEXURE FORMULA 291 6.4 THE FLEXURE FORMULA 2 9 1 6.4F THE FFLEXURE FORMULA 6.4 THE LEXURE ORMULA 6.4 THE FLEXURE FORMULA 291 6.4 THE FLEXURE FORMULA 6.4 THE FLEXURE FORMULA 291 EXAMPLE 6.13 EXAMPLE 6.13 6.13 EXAMPLE 6.13 EXAMPLE 6.13 6.13 ! EXAMPLE Şekilde enkesiti verilen konsol kirişin a-a kesitinde oluşacak maksimum eğilme gerilmesini EXAMPLE 6.13hesaplayınız. The beam shown in Fig. 6–27a2.6has kN a cross-sectional area in the shape The beam shown in Fig. 6–27a has a cross-sectional area in the shape 291 2 96.4 1 THE FLEXU 291 2.6 kN 2.62.6 kNkN The shown in in Fig. 6–27a hashas a cross-sectional area in in thethe shape beam shown Fig. a cross-sectional area shape 2.6 kN in the wn in Fig.The 6–27a hasThe abeam cross-sectional area the shape 2.6 kN beam shown in Fig. 6–27a has 6–27a aincross-sectional area shape of a channel, Fig. 6–27b. Determine the maximum bending stress that of a channel, Fig. 6–27b. Determine the maximum bending stress that 13 13 12 12 of of a channel, Fig.Fig. 6–27b. Determine the maximum bending stress that athe channel, 6–27b. Determine the maximum bending stress that 13 13 12 12 ig. 6–27b.ofDetermine maximum bending stress that 13 12 a channel, Fig. 6–27b. Determine the maximum bending stress that 13 12 5 5 occurs in5 the beam atin section a–a. occurs occurs in occurs the beam at section a–a. 5 kN 5 a 2.6 in the beam at section a–a. The beam shown in Fig. 6–27a has a cross-sectional area the shape in the beam at section a–a. a 5 a eaminatthe section occurs in the beam at section a–a. 2.6 kN a a rea shapea–a. of a channel, Fig. 6–27b. Determine the maximum bending stress that 13 12 SOLUTION SOLUTION nding stress that SOLUTION 13SOLUTION 12 5 occurs in the beam at section a–a. SOLUTION a 5 Internal Moment. Here the support reactions do not do have a beam’s Internal Moment. Here the beam’s 2support reactions1 m do not m 2m Internal Moment. Here the beam’s support reactions not have 2 m 1 m1have Internal Moment. Here the beam’s support reactions do not have 2 m m Internal Moment. Here thethe beam’s support reactions do not ma ment. Here thedetermined. beam’s support reactions do not have 2Instead, m have 1 m 2 mof sections,athe1segment to be Instead, by method of sections, the segment to the to be determined. by the method to the a SOLUTION a to be determined. Instead, byby thethe method of of sections, thethe segment to ato thethe (a) to be determined. method sections, toleft be Instead, byInstead, the ofto sections, the segment tosegment the (a)(a) ed. Instead, by determined. the method of sections, themethod segment the of left section a–a can be used, Fig. 6–27c. Inofparticular, note that (a) In particular, left section a–a can be used, Fig. 6–27c. note that (a) of section a–a can be used, Fig. 6–27c. In particular, note that Internal Moment. Here the beam’s support reactions do not have left of section a–a can be used, Fig. 6–27c. In particular, note that 2 m 1m 250 mm left of section a–a can be used, Fig. 6–27c. In particular, note that a–adocan used, Fig.internal 6–27c. In particular, note through that the resultant axial force N1passes the centroid ofaxial the force N passes 250 ons notbe have 250 mm 2 m m 250 mm the resultant internal through the centroid of the a mm the internal axial force Nthrough passes the centroid thethe _toof to the beresultant determined. Instead, by the method ofthrough sections, the segment resultant internal force N passes through the 250 mmof _ the resultant internal axial force N the centroid ofcentroid thebe aaxial 20 mm (a)moment _ mm cross section. Also,through realize that thepasses resultant internal moment must y ! 59.09 _ e segment to the nternal axial force N passes the centroid of the _ cross section. Also, realize that the resultant internal must be y ! 59.09 mm 20 mm (a) 20 mm cross section. Also, realize that the resultant internal moment must be y 20 mm ! 59.09 mm _ left of section a–a can be used, Fig. 6–27c. In particular, note that cross section. Also, realize that the resultant internal moment must be y A cross section. Also, that the resultant internal moment must be y ! 59.09 ! 59.09 mm N mm N 20 mm calculated therealize beam’s neutral axis at section cular,realize note that C a–a. Also, thatcalculated theabout resultant internal moment must be ya–a. 250 A mm 200Amm calculated about at Nsection ! 59.09 mmthe beam’s neutral A N axis N themm beam’s neutral ata–a. section a–a. the resultant internal axial force Naxis passes through the centroid of the C C calculated about the beam’s neutral axis at section a–a. calculated about theabout beam’s neutral axis ataxis, section 250 200 mm C A N To find the location of the neutral the cross-sectional area is 200 mm 200 mm _ of the utcentroid the beam’s neutral axis at the section a–a. To find the location of the neutral axis,mm the cross-sectional 15 15 mm area 20 mmis To find location of of the neutral axis, thethe cross-sectional area is mm 200 _cross section. Also, realize that the resultant moment be find the location the neutral axis, cross-sectional area isy ! 59.09 To find theTo location of the neutral axis, the cross-sectional area isCmust 15 mm mm subdivided into three composite parts shown in internal Fig.into 6–27b. Using 15as mm 20 mmas mm moment be neutral y ! 59.09 15 mm 15 mm 156–27b. mm A15 15 mm mm subdivided three composite parts shown in Fig. Using locationmust of the axis, the cross-sectional area is N subdivided into three composite parts as shown in Fig. 6–27b. Using calculated about the beam’s neutral axis at section a–a. subdivided into three composite parts as shown in Fig. 6–27b. Using subdivided into three composite parts as shown in Fig. 6–27b. Using 15 mm C 15 mm NAppendix A, we have A Eq. A–2 of (b) 200 mm Eq. A–2 of Appendix A, we have Cin A, o three composite parts asthe shown Fig. 6–27b. 200Using mm A–2 of Appendix wethe have (b)(b) Eq. A–2Eq. ofTo Appendix A, we have find location of neutral axis, the cross-sectional area is (b) Eq. A–2 of Appendix A, we have 15 mm ectionalA,area is©yA 15 mm pendix we have 2[0.100 m]10.200 m210.015 m2 + as [0.010 m]10.02 m210.250 m2m210.015 (b)Using 15 mm 15 mm parts subdivided into threem]10.200 composite shown in Fig. 6–27b. 2[0.100 m]10.200 m2 + [0.010 m]10.02 m210.250 m2 Çözüm: ©yA 2[0.100 m210.015 m2 + [0.010 m]10.02 m210.250 m2 y =©yA =©yA 2[0.100 m]10.200 m210.015 m2 + [0.010 m]10.02 m210.250 m2 ig. 6–27b. Using 2[0.100 m]10.200 m210.015 m2 + [0.010 m]10.02 m210.250 m2 ©yA = m210.015 m2y += 0.020 m2 210.200 m210.015 m2 + 0.020 m10.250 A,m]10.02 we havem210.250 (b) yEq. =y=A–2 =Appendix y = ©A [0.100 m]10.200 m210.015 [0.010 m2+ m10.250 =©Aofm2 =+ 210.200 ©A m2 (b) 210.200 m210.015 m2 0.020 m10.250 m2m2 ©A 210.200 m210.015 m2 + 0.020 m2 ©A 210.200 m210.015 m2m10.250 + 0.020 m10.250 = 0.05909 m = 59.09 mm m2 + [0.010 m]10.02 ©yA m2 2[0.100 210.200 m210.015 + 0.020m]10.200 m10.250m210.015 m2 = 0.05909 m =m210.250 59.09 mmm2 m]10.02 m210.250 == 0.05909 mm =mm 59.09 mm m 0.05909 = 59.09 ym2 == 0.05909 = = 59.09 mm This dimension 6–27c. m210.015 m2 + 0.020 m10.250 m22.4 kN 6 2.4 kN .05909 mm2= 59.09 mm©Ais shown in Fig.210.200 m10.250 This dimension is shown in2.4 Fig. 6–27c. ThisApplying dimension is moment shownisinshown Fig. 6–27c. This dimension in in Fig. 6–27c. kN 2.42.4 kNkN V the ofFig. equilibrium about the neutral 6 This dimension isequation shown 6–27c. 66 0.05909 m 1.0 kN Applying the moment equation of equilibrium about the neutral = 0.05909 m = 59.09 mm Vm M V V Applying the moment equation of equilibrium about neutral Applying thethe moment equation of of equilibrium about thethe neutral 1.0 kN 0.0 n is shown axis, in Fig. we6–27c. have 2.4 kN the 0.05909 m0.05909 Applying moment equation equilibrium about neutral 1.0 kN 6 1.0 kN N M 1.0 kN 0.05909Mm axis, we have M axis, we axis, have wewe have is shown V 2.4 kN e moment equation of equilibrium about the neutral Fig. 6–27c. N axis, NN 0.05909 m 6 1.0M kN = 0 d + ©MThis =dimension 0; have 2.4 kN12 m2 +in1.0 kN10.05909 m2 2.4 C NAkN M 6 d + ©M = 0; 2.4 kN12 m2 + 1.0 kN10.05909 m2 M = 0 V 2 m Applying the moment equation of equilibrium about the neutral NA 0.05909 mC d + ©MNA = 0; 2.4 kN12 m2 + 1.0 kN10.05909 m2 M = 0 d +d+ ©M = 0; 2.4 kN12 m2 + 1.0 kN10.05909 m2 M = 0 N C 1.0 kN # V NA out the neutral 2 ©MNA0.05909 = 0; M = kN12 4.859m2 kN +m1.0 kN10.05909 m2 - M = 0 M m2.4 C axis,1.0 wekN have M # m (c) 2 m2 m M = 4.859 kN2# m N # m kN Mm2 = 4.859 =kN ; 2.4 kN12 m2 + 1.0 kN10.05909 -M MM = 4.859 0N4.859 (c) (c)(c) Section Property. The moment of about axis is C =inertia kNm# mthe neutral2 m Fig. 6–27the neutral axis is Section Property. The about d+ ©M = The 0;parallel-axis 2.4 kN12 m2 + 1.0 kN10.05909 m2of -the M =is0moment # mthe Property. moment of inertia about the neutral axis C NA determined using theorem toabout each three Section Property. The moment ofapplied inertia the neutral is isof inertia M = 4.859 kN - M = 0 Section (c) axis Section Property. of inertia about the neutral axis C The momentdetermined Fig. 6–27 2 mFig. 6–27 using the parallel-axis theorem applied to Fig. each of the three # 6–27 2parallel-axis mcross-sectional determined using theusing theorem applied to each of the three composite parts of the area. Working in meters, we have m M = 4.859 kN determined the parallel-axis theorem applied to each of the three the theorem applied to the each of the three area. Working(c) erty. The momentdetermined of inertia using about theparallel-axis neutral axis is composite parts of cross-sectional in meters, we have (c) composite parts of parts the cross-sectional area. Working in meters, we have Fig.we 6–27 composite of of thethe cross-sectional Working meters, have 1Section Property. The moment of area. inertia about in thein neutral axis is composite parts cross-sectional area. Working meters, we have the parallel-axis theorem applied to of the three 3 each 2 engneutral axis is Fig. 6–27 I = c1 10.250 m210.020 m2 + 10.250 m210.020 m210.05909 m - 0.010 m2 d 1 3 2 2 Fig. 6–27 determined using the parallel-axis tom210.020 each of the three 1area. s of the cross-sectional wetheorem have I = m210.05909 c applied 10.250 m2+ 10.250 3 m210.020 2m210.020 m210.05909 m - 0.010 m2 d 1 Working I three = c 12 10.250 m210.020 m23 in + meters, 10.250 m 0.010 m2 d each of the I = c 10.250 m210.020 m2 + m210.05909 m 0.010 m2 d 3 10.250 m210.020 2 12 I = c 10.250 m210.020 m2 + 10.250 m210.020 m210.05909 m 0.010 m2 d 12composite parts of the cross-sectional area. Working in meters, we have 12 1 meters, we have +32+ c1 10.250 10.01512 m210.200 m23 + 10.015 m210.100 m - 0.05909 m22 d 1 2d 0 m210.020 m2 mm210.200 - 0.010 m2 3 2 11 m210.020 m210.05909 12 3 3 22m210.200 m210.100 m - 0.05909 m22 d + 2 c 10.015 m210.200 m2-m2 + d10.015 3 + 2 c I 10.015 m210.200 m2 + 10.015 m210.200 m210.100 m - 0.05909 1 10.250 m210.020 m2 ++3 10.250 m210.020 m210.05909 0.010 m2 dd 2 += 22+cdc -6 10.015 m210.200 m2m2 10.015 m210.200 m210.100 mm -0.05909 m2m2 12 4 12 0.05909 m - 0.010 m2 2c 10.015 m210.200 + 10.015 m210.200 m210.100 m 0.05909 d m = 42.2611012 12212 -6 4 -6 4 3 2 2 m = 42.26110 -6 21m 42.26110 5 m210.200=m2 + =10.015 m210.200 m210.100 m - 0.05909 m2 d 2-6m24m 42.26110 4 Maximum Bending bending stress occurs = + 2c 10.015Stress. m210.200The m23maximum + 10.015 m210.200 m210.100 m -at0.05909 m22 d 2 42.26110 0.100 m - points 0.05909 m2 d 12 Maximum Bending Stress. The maximum bending stress occurs at farthest away from the neutral axis.maximum This is at the bottom of the Maximum Bending Stress. The maximum bending stress occurs at 6 Maximum Bending Stress. The bending stress occurs at 2 m4 -6 4 Maximum Bending Stress. The maximum bending stress occurs points farthest away from the neutral beam, Thus, c = 0.200 m 0.05909 m = 0.1409 m. points farthest away from neutral axis. This is at theisbottom of the of the at axis. This is at the bottom of the 2 m thefrom = 42.26110 points farthest away the neutral axis. This at the bottom points farthest awaymfrom the neutral axis. at -the bottom beam, c = This 0.200is m 0.05909 m of= the 0.1409 m. Thus, 3 stress beam, c beam, Thus, = 0.200 - 0.05909 = m. nding Stress. The maximum bending at 4.859(10 ) 0.1409 N #m m10.1409 m2 c =mMc 0.200 m - 0.05909 =occurs 0.1409 m. Thus, Maximum Bending Stress. The maximum bending stress occurs at beam, Thus, c = 0.200 m 0.05909 m = 0.1409 m. 3 smax =axis. This Ans. =4.859(10 3the bottom # m10.1409 ngaway stress occurs from the at neutral is at42.26110 the = 16.2 MPa -6 4 m2 4.859(10 ) N # m10.1409 m2 ) Nthe 3 2 m#of McI away points=farthest neutral axis. =This is=MPa atMc the=bottom of the 4.859(10 )N m2 3 m10.1409 Mc from # s = 16.2 MPa Ans. Ans. = Thus, 16.2 the of the max 4.859(10 ) N m10.1409 m2 00 m bottom - 0.05909 msmax = 0.1409 m. -6 4 -6 4 Ans. = m Mc =42.26110 = I16.2 MPa 42.26110 2 m beam, Thus, cs =Imax 0.200 0.05909 m = 0.1409 m. -6 4 Show that at the top of the beam the bending stress is s¿ = 6.79 MPa. s Ans. 2 m = = = 16.2 MPa max I 42.26110 2 m -6 4 3 I # m10.1409 42.26110 2 is ms¿ 4.859(10 Nthe m2 Mc Show #bending that)at top offorce the beam the bending stress =V 6.79 MPa. that atm2 the ofkN thewill beam the bending stress is s¿ = 6.79 MPa. The normal of N =beam 1MPa kN3)the and shear force =top 2.4 4.859(10 NShow m10.1409 Mc that at the top of the stress is s¿ = MPa 6.79 MPa. Ans. = = NOTE:Show = 16.2 -6 4 =the top sadditional Ans. =Nof=the = 16.2 Show that at beam the bending stress is s¿ =will 6.79 MPa. max also contribute stress on the cross section. The superposition I MPaNOTE: -6 4 42.26110 2 m The normal force of 1 kN and shear force V = 2.4 kN NOTE: The normal force of N = 1 kN and shear force V = 2.4 kN will Ans. 16.2 Bu çözümde, a-aI kesitinde varolan normal kuvvet ve kesme kuvvetinin katkıları dikkate 42.26110 2and m shear NOTE: The normal force of N = 1 kN force V = 2.4 kN will of all these effects will be discussed in Chapter 8. also contribute additional stress on the cross section. The superposition NOTE: The normal force of N = 1 kN and shear force V = 2.4 kN will also contribute additional stress on the cross section. The superposition alınmamıştır. 8. bölümde etkiler alınacaktır. he top of the beam the bending stress s¿ stress =bu6.79 MPa. also contribute additional on theele cross section. Show that at the ofis the beam the stress is The s¿ =superposition 6.79 MPa. in Chapter 8. ofMPa. all these effects will betop discussed instress Chapter 8. these also contribute additional on the cross section. The superposition ofbending all effects will be discussed s¿ = 6.79 all1these effects will be discussed Chapter 8. force V = 2.4 kN will normal force of Nof =of kN and shear force =N 2.4 kN NOTE: The normal force = 1in kN and shear all these effects will beVof discussed inwill Chapter 8. V = 2.4 kN will eceadditional stress on the cross section. The superposition also contribute additional stress on the cross section. The superposition The superposition ects will be discussed Chapter 8. will be discussed in Chapter 8. of allin these effects PAGE !83 6.5 Unsymmetric Bending 6.5 Unsymmetric Bending 6.5 Unsymmetric Bending When developing the flexure formula weflexure imposed a condition that a condition that When formula we imposed 302 C Hdeveloping APTER 6 B Ethe NDING the cross-sectional area be about anwe axisimposed perpendicular the When developing thesymmetric flexure formula a condition that the cross-sectional area be symmetric about an to axis perpendicular to the neutral axis; furthermore, the resultant internal moment M acts along the cross-sectional area be symmetric about an axis perpendicular to the neutral axis; furthermore, the resultant internal moment M acts along Axis of symmetry Axis of symmetry axis. Such is neutral the caseaxis. for Such the “T” orinternal channel sections shown neutral axis; furthermore, the resultant Mchannel actsinalong the is the case formoment the “T” or sections shown in Axis of symmetry the neutral Unsymmetric 6–29. These conditions, are unnecessary, and insections this section the axis. Such ishowever, the caseconditions, for the “T” or channel shown in in this Bendin 6.5 SimetrikC HOlmayan Fig. 6–29. These however, are unnecessary, and section 302 A P T E R 6 Fig. BEğilme EN D I N Gneutral we will the flexure formula can also be applied either to aapplied either to a Fig.show 6–29.that These however, are unnecessary, and in this weconditions, will show that the flexure formula can also besection Neutral axis Neutral y area When developing formula beamwe having ashow cross-sectional of any shape or tobe a beam having atotoaflexure will axis that flexure formula can also applied either abeam having beamthe having a cross-sectional area of any shape orthe a w Neutral axis the cross-sectional area be symmetric abou resultant internal that acts in anyofdirection. beam havingmoment aresultant cross-sectional area any shape to adirection. beam having a internal moment that acts or in any Unsymmetric Bending neutral axis; furthermore, the resultant in M x Axis of symmetry resultant internal moment that acts in any direction. M x z z Moment Applied About Applied PrincipalAbout Axis. Principal Consider beam’s the neutralthe axis. Such is the case the “T” Moment Axis. Consider thefor beam’s M x z y developing the flexure formula we imposed a condition that cross Moment section to When have the unsymmetrical shape shown in 6–29. Fig. 6–30a. As in Applied About Principal Axis. Consider the beam’s Fig. These conditions, cross section to have the unsymmetrical shape shown in Fig. however, 6–30a. Asare in un thetocross-sectional area be symmetric about axis perpendicular toformula the y y Sec. 6.4, thesection right-handed x,the y,unsymmetrical zright-handed coordinate system is established such cross have the shape shown inan Fig. 6–30a. As in will show that the flexure Sec. 6.4, x, y, zwe coordinate system is established such can axis axis; furthermore, resultant moment M acts along y Axis ofthat symmetry the isneutral located at the centroid Cthe onat the cross section, the Sec.origin 6.4, the right-handed x, y, z Neutral coordinate is C established such beaminternal having athe cross-sectional that the origin is located thesystem centroid onand cross section,area andof theany the neutral axis. Such is the case for the “T” or channel sections shown resultant moment M internal acts along the + zM require theaxis. stress thatinternal the origin is located at the centroid Caxis. on the cross section, and resultant internal moment that actsin in any d resultant moment actsWe along the +z We the require the stress Axis of symmetry Axis of symmetry Fig. over 6–29. These conditions, however, are We unnecessary, and in this section M distribution acting the entire cross-sectional to require have a the zero resultant internal moment M acts + z area axis. stress x along distribution acting over the entire cross-sectional area to have a zero z Axis of symmetry we will show that the flexure formula can also be applied either to a Moment Applied About force resultant, the resultant internal moment about the y axis to be zero, distribution acting over the entire cross-sectional area have a zero force resultant, the resultant internal moment about the y axis toPrincipal be zero, Neutral axis beam having a cross-sectional area of any shape or to a beam having a cross section to have the unsymmetrical and the resultant internal moment about the moment z axis to about equal M.* force resultant, the the resultant internal moment about the ythe axis to beto zero, and resultant internal zThese axis equal M.* These sha y moment that acts in resultant internal anymathematically direction. 6.4, the right-handed x, y, z coordina conditions be expressed mathematically by considering the the resultant internal moment the zSec. axis to equal M.* These Neutral axis three and three conditions can about be expressed by considering the Neutral axiscan M x that the origin is located atz),the centroid C z acting on the differential element dA located at (0, y, z), Fig. 6–30a. conditions can be expressed mathematically by considering the Neutralforce axis three force acting on the differential element dA located at (0, y, Fig. 6–30a. Asal eksenlere göre döndüren moment etkisi Moment Applied About Principal Axis. Consider the resultant internal moment M beam’s acts along the This force is dF = s dA, and therefore we have force acting on the differential element dA located at (0, y, z), Fig. 6–30a. This force is dF = s dA, and therefore we have M M symmetry z x cross section to Axis haveofthe unsymmetrical shape shown in over Fig. 6–30a. As incross-s zx 6 distribution acting the entire This force is dF = s dA, and therefore we have M y x z Sec. 6.4, the right-handed x, y, z coordinate system is established such mome 6 force resultant, the resultant internal FR = ©Fx ; that the F 0©F = located -; s dA (6–14) = 0 = s dA (6–14) R x origin is at the centroid C on the cross section, and the Fig. 6–29 Fig. 6–29 and L the about t A resultant internal FR = resultant ©Fx ; 0L =A - Msacts dAalong (6–14) moment internal moment the + z axis. We require the stress Fig. 6–29 Neutral LA axis three conditions can be expressed mathe Axis of symmetry overzsthe areadifferential to have aelement zero 1MR2y = ©My ;distribution 0©M = dAentire 1MR2y =acting 0 =cross-sectional - acting zs dAon(6–15) (6–15)dA y; force the L A L A force resultant, the resultant internal moment about the y axis to be zero, 1MR2y = ©My ; 0 = - zs dA This force is dF = s (6–15) dA, and therefore we x L z the resultantMinternal A 6 and moment about the z axis to equal M.* These 1MR2z = ©Mz ; = z ; ys dA 1MR2z =M©M M = ys dA (6–16) (6–16) mathematically considering the Neutral axis three conditions can LA=be expressed LAFR = ©Fx ; by(6–16) 0 = - s dA 1MR2z = ©Mz ; M ys dA force actingFig. on6–29 the differential element dA located at (0, y, z), Fig. 6–30a.LA LA This force is dF = s dA, and therefore we have M x z 6 y R2y = ©My ; 1M 0 = - zs d y y y LA y s smax (6–14) F y= ©Fx ; 0 = - s dAmax z dFR! z dF ! sdA Fig. 6–29 sdA L A smax303 6.5 UNSYMMETRIC BENDING 1M ys dA s M = R2zs = ©M z; z dF ! sdA cs L A c dA 1M 2 = ©M ; dA M M y 0 = - zs dA y (6–15) R y y c 6.5 L UANSYMMETRIC ByENDINGx M 303 x As shown in Sec. 6.4, Eq. 6–14 is satisfied since the 6.5 z dA axisUNSYMMETRIC passes BENDING 303 6.5 UxNSYMMETRIC BENDING through the centroid of the area. Also, since the z axis represents the C C 1MR2z = M ©M MM= ys dA (6–16) y y y z; neutral axis for the As crossshown section, stress willis vary linearly from C L A in the Sec.normal 6.4, Eq. 6–14 satisfied since the z axis passes x x z y M z z dF distribution Bending-stress Bending-stress ! sdA nzero Sec.at6.4, 6–14axis, is satisfied since the axis passes yzshown = c, Fig. the Eq. neutral Hence As in 6–30b. Sec. 6.4, Eq. the 6–14 isx satisfiedthe since the zdistribution axis passes through to thea maximum centroid ofat the area. Also, (profile view) z since the z axis represents (profile view) Bending-stress distribution entroid of the area.isAlso, since the z -axis represents the s = (y>c)s . stress distribution defined by When this equation is through the of thewill area. Also, since the z axis(b) represents the max neutral axis for the cross section, thecentroid normal vary linearly (a)stress (a) from (profile view) dA y (b) the cross section, the6–16 normal stress willneutral vary linearly from substituted into Eq. and integrated, it leads to the flexure formula y axis for the cross section, the normal stress will vary linearly from c, Fig. 6–30b. Hence the zero at the neutral axis, to a maximum at y = (a) (b) smax y = c, Fig. utral to a maximum at 6–30b. Hence the smax axis, = Mc>I. When it is substituted into Eq. 6–15, we get Fig. 6–30 y = c, zero at the neutral axis, to a maximum at Fig. 6–30b. Hence the Fig. 6–30 z stress distribution is defined by s = -(y>c)s max . When this equation dF ! sdAis C ion is defined by s = -(y>c)s max . When this equation is Fig. 6–30 s s flexure = -(y>c)s stress distribution is max . When this equation y M substituted into Eq. 6–16 and integrated,isitdefined leads toby the formula c x o Eq. 6–16 and integrated, it leads to the flexure formula -smax dA substituted into Eq. 6–16 and integrated, it leads to the flexure formula M z y smax = Mc>I. When it isyzsubstituted intothat Eq.moments 6–15, get thethat dA = 6–15, *The condition about y axis be equal to zero in was not considered in B *Thewe condition moments about the ywas axisnot be considered equal to zero When it is substituted into0Eq. sget = 6.4, Mc>I. When it is substituted into Eq. 6–15, we get c we x maxSec. LA since the bending-stress distribution was symmetric with respect to the y axis and such 0 = which requires y y y 6.5 6.5 Sec. 6.4, since the bending-stress symmetric respect to *The condition that moments about the y axisdistribution be equal towas zero was notwith considered inthe y axis and such (a) a distribution stress produces moment about therespect yzero axis.moment See Fig. 6–24c. a distribution ofCstresszero automatically produces thesuch y axis. See Fig. 6–24c. -s Sec. 6.4,ofsince theautomatically bending-stress distribution was symmetric with to the y about axis and max -smax y -smax M yz of dA = a distribution stress automatically produces zero moment about the y axis. See Fig. 6–24c. Bunu 6.15 denkleminde0 yerine yz dA c koyarsak yz dA x 0z = A L Fig. 6–30 c LA Bending-stress distribution c LA (profile view) (a) (b) Buradan, which requires yz dA = 0 *The condition that moments about the y axis be which requires LA Sec. 6.4, since the bending-stress distribution was symme Fig. 6–30 a distribution of stress automatically produces zero mo yz dA = 0 This integral isBu called the of inertia for the area. Asdenilmektedir. indicated yz dA = 0 product yz dA =y ve 0 z eksenleri kesitin asal integrale çarpım atalet L A momenti L A L A that moments about the y axis be equal to zero was not considered in *The condition in Appendix A, eksenleri it will indeed be zero provided the y and z axes are olarak seçildiğinde çarpım Sec. atalet momenti sıfır olacaktır. 6.4, since the bending-stress distribution was symmetric with respect to the y axis and such chosen as principal axes of inertia for the area. For an arbitrarily shaped Kesitin asal eksenlerinin EkaA da atalet dönüşüm denklemleri This integral is called theyönelimi product of inertia forverilen theofarea. As indicated stress automatically produces zero momentile about the y axis. See Fig. 6–24c. area, the the orientation of inertia the principal axes can always be distribution determined, is called product of for the area. indicated ThisAs integral is called the product of inertia for the area. As indicated in Appendix A, Kesitte it will indeed be zero provided the y eksenlerin and z axes yönelimini are hesaplanabilir. bir simetri ekseni varsa asal belirlemek kolay either the inertia transformation equations Mohr’s of inertia A,using it will indeed be zero provided the and zoraxes iny Appendix A,are it circle will indeed be zero provided the y and z axes are chosen as principal axes of inertia for the area. For an arbitrarily shaped olacaktır daima simetri eksenleri boyunca ve for buna yöneleceklerdir. as explained in Appendix A, Secs. A.4 and A.5. Ifshaped the area hasofan axis 6 cipal axes of inertia for the çünkü area. For anchosen arbitrarily as principal axes inertia thedik area. For an arbitrarily shaped area, thethe orientation ofaxes the can principal can always be determined, of symmetry, however, easilyaxes be of established ntation of the principal axes principal can always be the determined, area, orientation the principal axes can always be determined, using either the inertiaalong transformation equations or Mohr’s of inertia since will always be oriented the ofaxis of symmetry and circle inertiathey transformation equations or Mohr’s circle inertia using either the inertia transformation equations or Mohr’s circle of inertia as explained in Appendix A, Secs. A.4 and A.5. If the area has an axis 6 it. A.4 and A.5. If the nperpendicular Appendix A, to Secs. has aninaxis asarea explained Appendix A, Secs. A.4 and A.5. If the area 6 has an axis of symmetry, however, the principal axes can easily be established For example, consideraxes the members in Fig. 6–31. Inthe each of however, the principal can easily be established ofshown symmetry, however, principal axes can easily be established since they willthe always be oriented alongfor thethe axis of symmetry and these cases, y and z must define principal axes of inertia cross they willand always be oriented along the axis of symmetry and l always be oriented along the axis since of symmetry it. through 6–16. In Fig. 6–31a the section in order perpendicular to satisfy Eqs.to6–14 perpendicular to it. to it. For example, consider the members shown in Fig. 6–31. In each of located by symmetry, Figs. 6–31b and 6–31c Forinexample, consider thetheir members shown in Fig. 6–31. In each of e,principal consideraxes theare members shown in Fig.and 6–31. In each of these cases, y and z must define the principal axes of inertia for the cross orientation is determined using theofmethods of Appendix A. Since M !8the is4 principal PAGE these cases, and z must define axes of inertia for the cross nd z must define the principal axes inertia for they cross section in order to satisfy Eqs. 6–14 through 6–16. In Fig. 6–31a the applied aboutEqs. one 6–14 of thethrough principal6–16. axessection (z Fig. axis),in the stress order satisfy Eqs.is6–14 through 6–16. In Fig. 6–31a the er to satisfy In 6–31a thetodistribution principal axes are located by and in shown Figs. , and determined the flexure s and = symmetry, -6–31c My>I is for and 6–31c principal axes their located by 6–31b symmetry, and intheir Figs. 6–31b and 6–31c their are located byfrom symmetry, and in formula, Figs. 6–31b zare is determined using the of Appendix Since M each case. using orientation orientation is methods determined using the A. methods of is Appendix A. Since M is determined the methods of Appendix A. Since M is of symmetry, theaxis principal axes can easily be established 6 ndix A, Secs. A.4 and A.5. If the however, area has an sincecan theyeasily will be always be oriented along the axis of symmetry and r, the principal axes established perpendicular it. s be oriented along the axis oftosymmetry and For example, consider the members shown in Fig. 6–31. In each of these cases, y and z must define er the members shown in Fig. 6–31. In each of the principal axes of inertia for the cross section satisfy Eqs. 6–14 through 6–16. In Fig. 6–31a the st define the principal axesinoforder inertiatofor the cross 304 CHAPTER 6 BENDING principal6–16. axes In areFig. located by the symmetry, and in Figs. 6–31b and 6–31c their M tisfy Eqs. 6–14 through 6–31a Aşağıdaki kesitlerde asalusing eksenlere göreofdöndüren orientation is determined the methods Appendix A.moment Since M isetkisi ed by symmetry, and in Figs. 6–31b and 6–31c their u y applied about oneA. ofdağılımı the principal axesdenklemini (z axis), thekullanarak stress distribution is olduğunda gerilme eğilme belirlenebilir. ned using the methods of Appendix Since M is z determined from distribution the flexure is formula, s = - My>Iz, and is shown for he principal axes (z axis), the stress flexure formula, seach = -case. My>Iz, and is shown for Moment Arb y My, Mz = the resultant internal moment components directed along the principal y and z axes. They are positive if directed along the + y and + z axes, otherwise they are negative. Or, stated another way, My = M sin u and Mz = M cos u, where u is measured positive from the + z axis toward the 6 z z " " My, Mz = the resultant internal moment components directed along Fig. 6–32 y (c) My ! M sin u x Here, z s = the normal stress at the point Iz (b) Mz ! M cos u y (a) u M y, z = the coordinates of the point measured from x, y, z axes having their origin at the centroid of the cross-sectional area and forming a right-handed coordinate system The x axis is directed outward from the cross-section and the y and z axes represent respectively the principal axes of minimum and maximum moment of inertia for the area (6–17) Iy Myz + Mzy s = - x y x BENDING " " ! 304 " ! CHAPTER 6 ! ! ! Moment Arbitrarily Applied. Sometimes a member may be loaded such that M does not act about one of the principal axes of the cross section. When this occurs, the moment should first be resolved into components directed along the principal axes, then the flexure formula can be used to determine the normal stress caused by each moment component. Finally, using the principle of superposition, the resultant normal stress at the point can be determined. To show this, consider the beam to have a rectangular cross section and to be subjected to the moment M, Fig. 6–32a. Here M makes an angle u with the principal z axis. We will assume u is positive when it is directed from the + z axis toward the + y axis, as shown. Resolving M into components along the z and y axes, we have Mz = M cos u and My = M sin u, as shown in Figs. 6–32b and 6–32c. The normalstress distributions that produce M and its components Mz and My are shown in Figs. 6–32d, 6–32e, and 6–32f, where it is assumed that 1sx2max 7 1s¿x2max . By inspection, the maximum tensile and compressive stresses [1sx2max + 1s¿ x2max] occur at two opposite corners of the cross section, Fig. 6–32d. Applying the flexure formula to each moment component in Figs. 6–32b and 6–32c, and adding the results algebraically, the resultant normal stress at any point on the cross section, Fig. 6–32d, is then loaded such that cross section. W into component formula can be moment compon resultant normal sözkonusu To show this, Momentx Arbitrarily Applied and to be subjec loaded such that M does not act ab angle u with the cross section. When this occurs, th directed from th into (a) components directed along th M into compon formula can be used to determine My = M sin and y 304y CHAPTER 6 BENDING moment component. Finally, using stress distributi M y resultant normal stress at the point y My are shown c u y y To show this, consider the beam Moment Arbitrarily Applied. Sometime y that 1s x2max 7 x z and to be subjected to the momen loaded such that M does not act compressive about one ofstre the CHAPTER 6 BENDING M withoccurs, the principal z axis. We cross section.angle Whenu this thethe moment sho of cross secti M directed from along the +z axis toward into components directed the principal M Applying the fa (a) M y M into components along the z ad as Moment Arbitrarily Applied. Sometimes aformula membercan may be be used to determine normal M and the 6–32c, and z M = M sin u, and as shown in Fi loaded such that M does not act aboutxone of the principal of the y Finally, using momentaxes component. the point principle at any on th x M z stress distributions the moment should first be normal resolved stress at the point that can beproduce determi z resultant x cross section. When this occurs, u x z Myconsider are shown in Figs. into components directed(b) along the principal then the flexure To show this, the beam to 6–32d, have a6–32 rec z (a) yaxes, (c) 6 x z x 1s¿x2maxM, . By that andcaused to be by subjected the 7moment Fig. inspe 6–32 formula can be each 1sto x2max (c) used to determine the normal stress (b) z Fig. 6–31 Mz ! Mucos u the [1swill + 1s¿u compressive with principal zstresses axis. We assume moment component. Finally, using the principle of angle superposition, the x2max M Fig. 6–31 +y axis, axis toward the +z cross section, Fig.the 6–32d. resultant normal stress at the point can be determined.directed fromofthe (a) (b) u M into components along z and y axes, we Applying thethe flexure formula to ea To show this,döndüren consider themoment beam to have a rectangular cross section Herhangi bir eksene göre etkisi 304 6 to BEN NG x C H A P T E Rand = Mand sin and M as shown in Figs. 6–32b alge and z 6–32c, and adding the results beD I subjected to the moment M, Fig. 6–32a. Here My makes anu, that on produce and its the crossMsection, Fig. angle u with the principal z axis. We will assume u is stress positivedistributions whenatitany is point Here, showny in Figs. 6–32d, 6–32e, and 6–32f, directed from the +z axis towardy the +y axis, as M shown. y are Resolving y Moment Arbitrarily Applied. Sometimes a member may be (a) 7 1s¿ that M1s M cos u x2max . By inspection, the m M into components along the z and y axes, we have 6 z x=2max loaded such that M does not act about xone of the principal axes of the [1sx2max + 1s¿ x2max compressive stresses occur s] = the at n and My = M sin u, as shown in Figs. 6–32b and 6–32c. The normalz cross section. When this occurs, the moment should first be resolved Mz ! Mcos u Mcz of the crossMsection, 6–32d. stress distributions that produce M and its components and MFig. ! M sin u z y, z = the into components directed along the principal axes, ythen the flexure s = Applying the flexure formula to each moment co My are shown in Figs. 6–32d, andto6–32f, where it normal is assumed Iz havin y (b) formula can 6–32e, be used determine the stress caused by each and 6–32c, and adding the results algebraically, the that 1sx2max 7 1s¿moment inspection, the maximum tensile and area x2max . By component. Finally, using the principle of superposition, the at" any pointcorners on the cross section, Fig. 6–32d, is The thenx M ! resultant [1sx2maxnormal + 1s¿ xstress 2max] occur compressive stresses two opposite at the at point can be determined. of the cross section, Fig. u the y To6–32d. show this, consider the beam to have a rectangular Here, cross section 6 the flexure formula to each moment component z in Figs. 6–32b y Applying of mi x z and to be subjected to the x makes an x moment M, Fig. 6–32a. Here M and 6–32c, and zadding the results algebraically, the resultant normal stress area u with angle it is Mz ! Mcos u the principal z axis. We will assume u is positive whenM s = the normalzystressMatyzthe p at any point on the cross section, Fig. 6–32d, is then + z = the r = - My, M directed from the +z axis toward the +y axis, as shown.s Resolving Iz Iythe (a) u y, z =Mthe=coordinates of (b) y ! Msin (c) have thepo p M cos u M into components along the zMand y axes, we z having their origin at the along M = M sin u, and as shown in Figs. 6–32b and 6–32c. The normaly x Fig. 6–32 area M and and forming a Or, right s stress distributions that z Mz ! Mcos u Myz produce M and its components Mzy The x axis is directed ou wher Here, M are shown in Figs. 6–32d, 6–32e, and 6–32f, where it is assumed y (6–17) s = - y + y y andand z axes represe Iy. By inspection, the maximum the +y a (b) that 1sx2max 7Iz1s¿x2max tensile z of minimum and maxim compressive stresses [1sx2max + 1s¿ x2maxxs] occur at two opposite corners Iy, Iz = the p = the normal stress at the point area of the cross section, M Fig.!6–32d. axes, Msin u y, z = the coordinates of the point measured y My, Mz = in theFigs. resultant Applying the flexure formula to each moment component 6–32b internal mo Here, having their origin at the centroid of t y (c) algebraically, the resultant normal the principal and 6–32c, and adding the results stress y and z axe area and forming a right-handed coor +y and +z axe along the at any point on the cross section, Fig. 6–32d, is then The x axis is directed outward from t s = the normal stress at the point Fig. 6–32 Or, stated another way, M the y and z axes represent respective My ! Msin u y, z = the coordinates of the point measured from x, y, z axes u where is measured poso 6 z of minimum and maximum moment having their origin at the centroidxof the cross-sectional x +y axis z Mz ! Mcos u area and forming a right-handed coordinate Myzarea Mzy system Iy, Iz = internal the principal moments of + (6–17) s = M , M resultant moment compon The x axis is directed outward from the cross-section and y z = the I I axes, respectively. See Ap z y (b) (c) principal y and z axes. They are po the y and z axes represent respectively the principalthe axes z along the +y and +z axes, otherwise t of minimum and maximum moment of inertia for the x Fig. 6–32 Or, stated another way, My = M sin u area Here, u where is measured positive from the y y, Mz = the resultant internal moment components directed along M +y axis (c) the principal y and z axes. They are positive if directed s = +z theaxes, normal stress atthey the are point , Iz = the principal moments of inertia calcula Iynegative. along the +y and otherwise Fig. 6–32 axes, respectively. See Appendix A Mz =measured My = M sinofuthe M cos u, from Or,u stated another way,coordinates andpoint My ! Msin y, z = the x, y, z axes where u is measured positive from theat+zthe axis towardofthe having their origin centroid the cross-sectional +y axis area and forming a right-handed coordinate system Theofx inertia axis is calculated directed outward from thez cross-section and Iy, Iz = the principal moments about the y and the y and z axes represent respectively the principal axes axes, respectively. See Appendix A z PAGE !85 of minimum and maximum moment of inertia for the x area (c) x y (b) z z 6 " " " " " " PAGE !86 My, Mz = the result the princi y, z = the coord having th area and The x axi the y and of minim area My ! M sin u x Mz ! M cos u y (a) x z M u y " CHAPTER 6 BENDING " " " 304 s = the norma Here, Moment Arbitra loaded such that M d cross section. When into components dir formula can be used moment component. resultant normal stre To show this, consi and to be subjected angle u with the prin directed from the + M into components and My = M sin u, stress distributions My are shown in Fi that 1sx2max 7 1s¿x2 compressive stresses of the cross section, F Applying the flexur and 6–32c, and adding at any point on the cr y ection. When this occurs, the moment should first be resolved [(sx)max " (s¿x)max] stress distributions that p (6–18)the a of the ion of the Neutral Axis. The angle the flexure mponents directed along principal axes, then My are shown in Figs. 6–3 N y candetermine be determined by applying 6–17by each as in canFig. be 6–32d used to the normal stress Eq. caused that 1sx2max 7 1s¿x2max . B ,t since by definition nousing normal acts on neutral A component. Finally, thestress principle of the superposition, the x compressive stresses [1sx2m z z This equation defines the neutral axis for the cross section. Since the venormal stress at the point can be determined. [(sx)max " (s¿x)max] nt a of the cross section, Fig. 6–3 slope of this line is tan a = y>z, then how this, consider the beam to have a rectangular cross section 6 (sx)maxApplying the flexure form [(sx)max ! (s¿x)max] I M y z be subjected to the moment M, Fig. 6–32a. Here M makes an z y = and 6–32c, and adding the re 6.5 U NSYMMETRIC ENDING 305 Tarafsız eksenin yönelimi: NSYMMETRIC BBENDING MzIU u is positive 305 with the principal z6.5 axis. We will assume when it is y (e)at any point on the cross sect Iz (d) NSYMMETRIC BENDING 305 d from the +z axis toward the +y axis, as shown. (6–19) tan a Resolving = tan u6.5 U6.5 UNSYMMETRIC BENDING 305 I y y M = M cos u along the z and y axes, we have hecomponents proper algebraic M cos u and My = M sin u, then y z ! BENDING 6 6.5 UNSYMMETRIC 305 = the M sin u, as shown in Figs. 6–32b and 6–32c. The normalcoordinates y and ynd x y y, z axes form a right-handed system, and the proper algebraic Mz and distributions that produce M and system, its components The x, y, z stress axes form a right-handed and the proper algebraic z he resulting he Mz ! Mcos 6.5u y UNSYMMETRIC 305 st be assigned to the moment components and the coordinates Iz 6.5 UNSYMMETRIC BENDING 3 0 5BENDING y shown in Figs. 6–32d, 6–32e, and 6–32f, where it is assumed The x, y, z axes form a right-handed system, and the proper algebraic gns must be assigned to the moment components and the coordinates egative. s Here it can be seen that unless I the angle u, defining the direction = I tan u z y = ¢ ≤ [(s ) ! (s ¿ ) ] [(s ) ! (s ¿ ) ] z y (s ) x max x max max x max plying this equation. When this is the case, the xresulting stress x max (s¿x)max Iinspection, (b) y shen 2maxapplying 7 1s¿x2this .[(s By the maximum tensile and signs must be assigned to the moment components and the coordinates equation. When this is the case, the resulting stress of the moment M, Fig. 6–32a, will not equal a, the angle defining the x max [(s ) " (s ¿ ) ] ) " (s ¿ ) ] 6.5 U NSYMMETRIC B ENDING max nsile ifait of is positive and it is negative.system, and the proper algebraic xx max xx max [(s ) ! (s¿x)max y] The y,maxwhen zcompressive axes aifright-handed angle the y + 1s¿ 2form ssive stresses occur at corners y, axes form ait right-handed and the algebraic applying this equation. When this is the case, the resulting stressx max [(s" illzbe tensile if[1s isx2x, positive and compressive iftwo itproper isopposite negative. of] the neutral axis, Fig. 6–32d. maxinclination xsystem, max ! x)max ! (s¿x)max] N N [(s ) " (s ¿ ) ] signs must be toifthe components and x maxthe coordinates x max plying Eq. t be section, assigned to the moment and the coordinates ross Fig. 6–32d. willassigned be components tensile it ismoment positive and if[(s itx)ismax negative. a ation of 6–17 the Neutral Axis. The angle ofcompressive the [(sx)max ! (s¿x)max] " (s¿x)max] y when applying this equation. When this is the case, the resulting stress sying AAresulting The x, y, z axes form a right-handed system, and the proper algebraic on the neutral a Orientation of the Neutral Axis. The angle of the N lying this equation. When this is the case, the stress the flexure formula to each moment component in Figs. 6–32b Here, xis in Fig. 6–32d can be determined by applying Eq. 6–17 x [(s ) " (s ¿ ) ] x max x max zz it for ythe xcoordinates z The components N of the on and defines theFig. neutral axis theifsigns cross section. Since the will be tensile if is positive and compressive if it is negative. must be assigned to the moment and a Orientation of the Neutral Axis. angle eutral axis in 6–32d can be determined by applying Eq. 6–17 [(s ) " (s ¿ ) ] Important Points [(s ) ! (s ¿ ) ] [(s ) " (s ¿ ) ] nsile if it is positive and compressive it is negative. 2c, adding the results algebraically, the resultant normal stress x xmax x max x max max neutral max [(s 0, since by definition no normal stress acts onx xmax the Ax)max ! (s¿x)max] aa N sith line is= the tan then a = y>z, x resulting when applying this equation. When is the [(s case, the neutral axisnormal in Fig.stress 6–32dacts can determined by applying Eq. 6–17 son 0, since by definition no onbethe neutral z )this oint cross section, Fig. 6–32d, is then (s¿x))maxA]" (s¿ )stress x)max "[(s have [(s ) " (s ¿ ] (s ¿ ) 6 x max x max ] [(s ) ! (s ¿ ) ] x max s = the normal stress x [(sxx)max ] )max x max x max x¿x)max max ! (s xz a acts Neutral The angle ofa(sthe aAxis. tion of Orientation the Neutral Axis. The the will be tensile if of it isno positive and compressive if itneutral is negative. sof =maxthe 0, since with byangle definition normal stress on the A xis. We have [(sx)ma [(s x)max " (s¿x)max] N x the coordinates a •in The flexure formula can applied only when bending occurs z neutral Fig. 6–32dby can be determined by applying Eq. N6–17 My ! Msin u y,¿ )z = xis in Fig. 6–32d canaxis beaxis. determined applying Eq.be6–17 [(s We have x)max ! (s¿x)max] I M [(s ) " (s ¿ ) [(s ) " (s ] x max x max] y z x max x max (d) (e) [(sneutral )max ! (s¿x)maxThe ] Athe Iabout axes that represent the principal inertia for (d)Orientation of the Neutral Axis. angle a of the a A s =y 0, stress acts on axes the xof having their orig z by 0, since by with definition nosince normal stress actsno onnormal the neutral (f) z Mdefinition = yIz x x (6–19) tanhave a = Mcross tan u N z [(sx)max ! (s¿x)max][(s ) z These y zIy= section. have their originz at the centroid and neutral axisaxes in Fig. 6–32d 6–17 axis. We ave area and formin M [(sx)max "Eq. (s¿x)max ] IM yIz can be determined (d) by applying x max " (s¿x)max] y zy MM Iyyz a a Fig. 6–32 (cont.) z ! z y =by axis ofdefinition symmetry, if there is one, s = 0,ansince withalong no normal stress acts on the neutral (d)and + (6–17) s = - are oriented The x axisxAis dir eksende sıfır olacağından) z MzIy [(sx)(Tarafsız y Iyaxis. to [(sgerilme max ! (s¿x)max] x)max ! (s¿x)max] IIzz u, then Mysin = M cos u and My = M perpendicular Weit.M have yIz (d) the y and z[(s axes x)ma a = My =z M sin u, then z y = ince Mz = M cos uyand z of minimum and •MIzIfIthe is M applied about some arbitrary axis, then the y the moment zI y direction [(sx)max ! (s¿x)max] be seen that unless Iz = defining the Since and = Mu,cos u(s M (d) y Mzangle (d) x (sx)xmax )max y = M sin u, then MyIz (s¿x)max area moment must be resolved intothe components along I z equal a, the angle defining z each of the y = ment M, Fig. 6–32a,ywill not tan u z = ¢ ≤ ,= M cos u and I (s ) MzIisy determined byx max stress at a point z axes, MM sin u, uthen My, Mz = the Since and M cos Muy ≤=z and M sinthe u, then y Fig. Iyprincipal of the neutral axis, 6–32d. z= =M (d) resultant int y = ¢ tan Iz superposition of the stress caused by each of the moment(c) (sx)max the principal y a Iy y = ¢ tan u ≤ z (sx)max components. Since and then M = M cos u M = M sin u, s6–18) = the normal stress at the point I along the +y an z y y Iz s section. section. Since Since the Iz Fig. 6–32 z tan u ≤ z they = (6–18) ¢ Or, stated anoth portant Points (s ) u ≤zz axes y = from x max z = the coordinates of the x, y, ¢ tan (sx)max Iy point measured Iy z ation defines the neutral axis for the cross section. Since the 6 u where is measu having their origin at the centroid (sx)max of the cross-sectional 6 Iz(s¿x)max x)maxfor the cross section. Since the (6–18) his defines the neutral (s axis y = ¢ tan uz ≤ z his equation line tan a = y>z, +y axis (sx)max areaisand forming athen right-handed coordinate system exure formula can beaThis applied only when bending occursaxis for the Icross 6 y z (sx)max Since the equation defines the neutral section. ope of this line is tan then = y>z, (e) from the cross-section and The x axis is directed outward 6Iy, Iz = the principal mo (6–18) (e) axes that represent the principal axes for then the (sx)max (f) slope offor this line isof taninertia a =principal y>z, (6–19) the y and z axes represent respectively the axes z tion defines the neutral axis the cross section. Since the axes, respectively (6–19) 6 z Thisaxes equation defines the at neutral axis for and the cross section. Since(e)the section. These have their the centroid (sx)max Iz origin ! of minimum and maximum moment of inertia for the his line is tan then a = y>z, Fig. 6–32 (cont.) (6–19) tan =symmetry, riented along an ofaxis if there is one, and slope thisaof line is tan ua! then = y>z, (e) y 6 (s )max Iy a = IzThis area 6 y uequation defines the neutral z the cross section.(sSince (6–19) axis forx ! tan tan x)max the ndicular to it. (e) I z I y slope of this line is tan then a = y>z, = thethe resultant internal components tan athe =along tan u y ! (6–19) zfining direction moment is applied aboutmoment arbitrary axis,directed then (e) Izsome (sx)max (s¿x)max if directedIy fining the direction the principal y and z axes. They are positive y (e) I ! z (s¿x)each (6–19) a = components tan u angle defining thetan into max nt must be resolved along of uthe (6–19) tan a = tan angle defining the I y they are negative. an bealong seen the that+y unless I+z angle u, direction Iyy the z =axes, Iy the pal axes, and theand stress at aotherwise point is defining determined by (s¿x)max (e) Iz ! y ! M M = M sin u = M cos u, Or, stated another way, and oment M, Fig. 6–32a, will not equal a, the angle defining the Here it can be seen that unless I the angle u, defining the direction = I y z z y position of the stress caused by each of the moment (s¿x)max (6–19) tan a = tan u y u is M, +z measured positive thethat axis toward Here it will can from be Iz = defining the direction Ithe on ofwhere the neutral axis, 6–32d. f the moment Fig.Fig. 6–32a, notseen equal a, unless the angle the u,Idefining y the angle y onents. (sx¿ )max ! axis moment M,defining Fig. 6–32a, will not equal a, the angle defining the nclination of the neutral 6–32d. n be+y seen that unless Izofaxis, the angle u, the direction = the Iy Fig. y (s¿x)max Here it can be seen that unless I the angle u, defining the direction = I zangle y defining inclination of calculated the neutral axis, Fig. oment Fig. 6–32a, will not equal a, the (s¿x)max principal moments of inertia about the 6–32d. y the and z (s¿x)max z = theM, of the moment M, (s Fig. mportant Points ¿x)max6–32a, will not equal a, the angle defining the n of axes, the neutral axis, Fig. 6–32d. respectively. See Appendix A it can be seen that unless Iz = Iy the angle u, defining the direction n bending occurs inclination of the neutral axis, Fig. 6–32d. (s¿x)max Important Points Here ¿x)max a, the angle defining the nofbending occurs of the moment M, Fig. 6–32a, will not(sequal inertia for the (f) Important Points (s¿x)max of inertia for and the can be applied only of the neutral flexure formula when bending occurs axis, Fig. 6–32d. (f) inclination the centroid mportant Points (s¿x)max 6–32 (cont.) he centroid and ut that and represent the axes of when inertiabending for the occurs • axes The flexure formula canprincipal be Fig. applied only here is one, (f) Important Points Fig. 6–32 (cont.) (s ¿ ) here is one, andthat s section. These axes have their origin at the centroid andfor the • The flexure formula can be applied only whenx max bending(f)occurs about axes represent the principal axes of inertia Fig. 6–32 (cont.) (sx¿ )max Important Points oriented along an axis of symmetry, if there is one, and flexure formula can be applied only when bending occurs about axes that represent the principal axes of inertia for the crossthen section. (f) ry axis, the These axes have their origin at the centroid and Fig. 6–32 (cont.) • The flexure formula can be applied only when bending occurs t axes that represent the principal axes of inertia for the pendicular to it. (s¿x)max cross section. These axes have their origin at the ry axis, then the oriented (f) centroid and longare each of the along an axis of symmetry, if there is one, and Fig. 6–32 (cont.) about axes that represent the principal axes of inertia for the she section. These axes have their origin at the centroid and (f) are oriented along an axis of symmetry, if there is one, and ong each of the perpendicular to it. about some arbitrary moment is by applied axis, then the can be applied determined • The flexure formula only 6–32 when (cont.)bending occurs cross section. These axes have their at the centroidFig. and oriented along an axis of perpendicular symmetry, if there one,origin to it. is determined ment must be by resolved into components along each ofand the of the about axes that represent for6–32 the(cont.) • If themoment moment is applied about some arbitrary axis, then the the principal axes of inertiaFig. (f) are oriented along an axis of symmetry,byif there is one, and endicular to it. of moment theaxes, moment cipal and stress a moment point isisdetermined • If atthe applied some arbitrary then thecentroid and cross section. These axes their axis, origin at the must the be resolved into components along about each of thehave perpendicular to it. arbitrary Fig. 6–32 (cont.) e moment is axes, applied some axis, the components erposition the stress caused each the moment along an by axis of symmetry, moment must beoforiented resolved into along eachifofthere the is one, and principalof andabout the stressby at a are point is then determined mentsuperposition must be • resolved into components along each the If of thethe moment iscaused applied some arbitrary the mponents. perpendicular to it.moment principal axes, and the stress at a axis, pointthen is determined by stress byabout each ofof the cipalcomponents. axes, and moment the stress at be a point isofdetermined by must resolved into components along each ofofthe superposition the stress caused by each the moment • If ofthethe moment is applied about some arbitrary axis, then the rposition of theprincipal stress caused by each axes, and the stress atmoment a point is determined by components. moment must be resolved into components along each of the ponents. superposition of the stress caused by and eachthe of stress the moment principal axes, at a point is determined by components. superposition of the stress caused by each of the moment components. C H A P T E R 6 306 BENDING CHAPTER 6 BENDING 306 CHAPTER 6 BENDING MPLE 6.15 EXAMPLE 6.15 6.5eğilme UNSYMMETRIC BENDING 307 ! Şekilde verilen dikdörtgen enkesite 12 kNm momenti EXAMPLE 6.15 gösterildiği gibiThe tesir etmektedir. Kesitin meydana gelecek normal rectangular cross section shownköşesinde in Fig. is subjected a 6–33agerilmeyi The herbir rectangular cross6–33a section shown in to Fig. is subjected to a The rectangular cross section shown in Fig. 6–33a is subjected to a # # ve tarafsız eksenin yönelimini belirleyiniz. bending moment of M bending normal stress m. Determine = 12 kNmoment of M =the Determine the normal stress m. 12 kN # bending of and Determine the normal stress M = 12 kNthem.orientation 4.95 and MPa developed at each corner of moment the section, specify ofspecify developed at each corner of the section, the orientation of developed at each corner of the section, and specify the orientation of x the neutral axis. the neutral axis. the neutral axis. A E 2.25 MPa SOLUTION SOLUTION 2.25 MPa D E 0.2 m SOLUTION B it is seen that Internal Moment Components. By inspection the y Internal Moment Components. By inspection it is seen that the y Moment Components. By inspection it is seen that the y 0.2 m D 5Internal and z axes represent the principal of inertia since they are axes of and z axesaxes represent the principal axes of inertia since they are axes of N and zsection. axes represent thezprincipal axes of4.95 inertia since they are axes of 3 cross 4is subjected The rectangular cross sectionB shown in Fig. to required a the MPa symmetry for6–33a the As we have established the z symmetry for cross section. As required we have established C M ! 12 kN"m symmetry forstress the cross section. As required weThe have established the z the z # m.asDetermine bending moment of M = 12 kN the normal axis the principal axis for maximum moment of inertia. axis as the principal axis for maximum moment of inertia. as for maximum moment of inertia. The The developed at each corner and specify axis the orientation 0.2 m axis moment is resolved into its the y andprincipal zof components, where 0.1 m of the section, C moment is resolved into its y and z components, where moment is resolved into its y and z components, where the neutral axis. z 0.1 m 4 y 4 #m My = - 112 kN # m2 = M - 9.60 kN 4 112 kN # m2 = -9.60 kN # m 5 My = -y = 112 kN 5 # m2 = -9.60 kN # m SOLUTION 5 3 3# m 112 kN # m2 = 7.20 3 kN Internal Moment Components. By inspectionM it zis=seen = kN 112 7.20 # m2kN=# m2 # mkN # m 5 that the yMz = Mz112 7.20=kN 5 and z axes represent the principal axes(a) of inertia since they are axes of 5 (b) symmetry for the cross section. As required we have established the z of inertia about Section Properties. The moments the y andofz inertia axes Section Properties. The moments the yz axes and z axes Fig. 6–33 Properties. The moments of inertia about about the y and axis as the principal axis for are maximum momentSection of are inertia. The Çözüm: are moment is resolved into its y and z components, where 1 6 -3 1 6 Iy = 10.4 m210.2 m23 I1= 0.2667110 2 m34 m23 = 0.2667110 -3 4 = 10.4 m210.2 4 y 12 Iyof= the10.4 m210.2 m2 = 0.2667110-32 m4M2!m12 kN"m # # Orientation of Neutral Axis. The location z neutral axis 12 M = 112 kN m2 = 9.60 kN m y 12 5 6–33b, can be established 1 by proportion.3Along A (NA), Fig. -3 1the edge I = 10.2 m210.4 m2 2 m43 m23 = 1.067110 -3 4 1= 1.067110 z 3 I = 10.2 m210.4 BC,=we 112 require Iz = z 10.212m210.4 m2 = 1.067110-32 m54 42 m M kN # m2 = 7.20 kN # m 12 z 12 3 5 2.25 MPa 4.95 MPa D E Bending Stress. = Thus,Bending Stress. Thus, 10.2y m - zz2 Section Properties. The moments of inertiaz about the and axes Thus, Bending Stress. u ! #53.1$ are M0.450 Mzy yz - 2.25z =M4.95z M z z y y zMyz s = + My + m a ! #79.4$ Iz Iy s = -s =zz- =+ 0.0625 1 3 Iy Iz -3I2 zm Iy4 Iy = 10.4 m210.2 m2 =# 0.2667110 a 3 3 12 7.20110 2 N m10.2 m2 9.60110 2# ND# m1 - 0.1neutral m2 3 axis 3 3 In the same manner this is also the 7.20110 distance from to the # 3 2 N m10.2 m2 -9.60110 2 N m1-0.1 m2 s1B = + 2 N # m10.2 m2-3 -9.60110 = 22.25 MPa B m2Ans.C N # m1-0.1 4 7.20110 4 + in Fig. 6–33b. s = 2.25 MPa 3s -32 -3 4 0.2667110 + B 1.067110 m 2 m Ans. Ans. = = 2.25=MPa -3 4 Iz = 10.2 m210.4 m2 = B 1.067110 2 m 1.067110 4-32 m4 -3 0.2667110 1.067110-3of2 m 0.2667110 2 m4 2 m 12 can also establish We the orientation the NA using Eq. 6–19, 3 3 # # 7.20110 2 N m10.2 m2 - 9.60110 2# N m10.1 m2 3 3 3that #the 7.20110 2N m10.2 m2 2 N # m10.1 m2 which the angle+a axis makes with -9.60110 the N Ans. = 3-2z4.95 MPa sC =is-used to specify -9.60110 2 N m10.2 m2 Nor# m10.1 m2 -3 s4 7.20110 -3 4 = + = -4.95 1.067110 2 m 0.2667110 2 m Bending Stress. maximum Thus, C y -4.95 Ans. Ans. = + MPa MPa s principal axis. our-3sign4-3 convention, u must be 4 -3 4= C According to 1.067110 -3 4 2m 0.2667110 2m 1.067110 2 m 0.2667110 2 m 3 3 # # measured 7.20110 from the axis toward + y axis. comparison, in 3 2 N+ zm1 - 0.2 m2 the - 9.60110 2 NBym10.1 m2 3 (c) # m1-0.2 Myz 7.20110 2+ N m2 -9.60110 2 N # m10.1 sD6–33c, = - u = - tan-1 43-3= -453.1° +32uN=# m1-0.2 = 3-22.25 MPa 7.20110 m2 -9.60110 N # m10.1 m2Ans.m2 Thus, Fig. (or 306.9° ). -3 4 s = + = -2.25 + 1.067110 2 m 0.2667110 2 m D Ans. Ans. sD = + = -2.25 MPa MPa -3 4-32 m4 -3 4-32 m4 Iy 1.067110 0.2667110 Iz 1.067110 2 UmNSYMMETRIC 0.2667110 2m 6.5 B ENDING 307 3 3 # # 2 Na m1 0.2 2 N m1 - 0.1 m2 7.20110 - 9.60110 = -m2 tanm2 u 7.20110 3 3 3 2 N # m1-0.2 2 N # m1-0.1 -9.60110 1032 N # m10.2 m2 sE- 9.60110 2 N #tan m1 - 0.1 Ans. m2 = +32 N # m1-0.2 =324.95 MPa Iy 47.20110 m2-3 m2 N # m1-0.1 m2 -9.60110 -3 4 + s = = 4.95 MPa Ans. + = 2.25 MPa 1.067110 2 m 0.2667110 2 m E Ans. Ans. -3 4+ -3 4 = 4.95 MPa -3 4 -3 sE 4= -3 4 -3 4 -3 4 1.067110 0.2667110 .067110 2 m 0.2667110 2 m 1.067110 2m 1.067110 2m 2m 0.2667110 2m 2m aresultant =m2 tan1distribution - 53.1°2 #tan MPa -3 normal-stress has been sketched using 1032 N # m10.2 m2 - 9.6011032 N The m10.1 0.2667110 2 m4The4.95 resultant normal-stress distribution been sketched Ans. + = 4.95 MPa The resultant normal-stress distribution has been using using x these Fig. 6–33b. Since superposition applies, the distribution ishas sketched -3 4 -3 values, 4 .067110 2 m 0.2667110 2 m Fig. 6–33b. Since applies, the distribution is A a = - 79.4° these these Ans.superposition values,values, Fig. 6–33b. Since superposition applies, the distribution is linear as shown. 3 # linear as shown. E 1032 N # m1 - 0.2 m2 This- 9.60110 2 N m10.1 m2 linear as shown. 2.25 MPa in Fig. 6–33c. Using value of z calculated + result is shown = - 2.25 MPatheAns. 4 2.25 MPa D one obtains 0.2 m -32 mE4 1.067110 0.2667110 2m above, verify, using -3 the geometry of the cross section, that B 3 3 the same answer. 10 m 2 N # m1 - 0.2 m2 D 5 - 9.60110 2 N # m1 - 0.1 m2 N = 4.95 MPa Ans. + 3 4 4.95 MPa 1.067110-32 m4 0.2667110-32 m4 z C M ! 12 kN"m The 0.2 m has been sketched using C resultant normal-stress distribution these values, Fig. 6–33b. Since superposition applies, the distribution is z linear as shown. (a) (b) Fig. 6–33 PAGE !87 of Neutral Axis. The location z of the neutral axis M ! 12 kN"m 6 Fig. 6–33 Fig. 6–33 (a) Orientation of Neutral The location z of the neutral Orientation of Neutral Axis. Axis. The location z of the neutral axis axis Fig. 6–33 Fig. 6–33b, be established by proportion. the edge (NA), (NA), Fig. 6–33b, can becan established by proportion. AlongAlong the edge we require BC, weBC, require (b) neutral axis ng the edge 2.25 MPa 4.95 MPa MPa 4.95 Orientation of Neutral2.25 Axis. The location E = MPaz of the neutral axis = z 10.2 m10.2 m - z2 Tarafsız eksenin yönelimi: z - z2 (NA), Fig. 6–33b, can be established by proportion. Along the edge 0.450 - = 2.25z = 4.95z BC, we require 0.450 2.25z 4.95z M ! 12 kN"m z = 0.0625 m 2.25 MPa z = 0.0625 4.95 MPa m E A = In the same manner this is also the distance from D to the neutral axis z 10.2 m z2 5 6 4 In the same manner this is also the distance from D to the neutral axis E neutral axis g Eq. 6–19, with the z or n, u must be mparison, in a B A A 5 ED M ! 12 M kN! "m12 kN"m 4 3 D 5 4 3 M ! 12 kN"m A u ! #53.1$ u5! #53.1$ 4 z 3 a ! #79.4$ D a ! #79.4$ a a B u ! #53.1$ C B C in Fig. 6–33b. in3 Fig. 6–33b. 0.450 - 2.25z = 4.95z z D We can also establish the orientation of the NA using Eq. 6–19, We can also establish the orientation of the NA using Eq. 6–19, a ! #79.4$ = angle 0.0625am which is used to specify zthe that the axis makes with the z or N is used to specify the angle a that the axis makes with the z or N a u which ! #53.1$ y maximum principal axis. According to our sign convention, u must be Inmaximum the same principal manner this isAccording also the distance D to the neutral axisbe y C axis. to our from sign convention, ucomparison, must B z measured from the + z axis toward the + y axis. By in (c) inmeasured Fig. 6–33b. from the +-ztan axis -1 4toward the + y axis. By comparison, in a! #79.4$ Fig. 6–33c, u =-1 (c) 3 = - 53.1° (or u = + 306.9°). Thus, 4 C N y (c) E Ans. z calculated one obtains (b) We6–33c, can also NA). using Thus, Eq. 6–19, Fig. u = establish - tan 3 the = - orientation 53.1° (or u =of+the 306.9° which is used to specify the angle Iaz that the axis makes with the z or N tanIza = tan u y maximum principaltan axis. a According = tan uIy to our sign convention, u must be measured from the +z axisIytoward the +y -3 axis. 4 By comparison, in 1.067110 2 m (c) a = (or-3u2 = tan1 - 53.1°2 Fig. 6–33c, u = -tan-1 43 =tan -53.1° ). Thus, 1.067110 m4+306.9° -3 4 0.2667110 tan1 2 m- 53.1°2 tan a = Iz0.2667110-32 m4 a =u - 79.4° Ans. tan a = tan Iy- 79.4° a = Ans. This result is shown in Fig. 6–33c. Using the value of z calculated -3 m4of the Thisabove, result verify, is shown in 1.067110 Fig.geometry 6–33c.2 Using the value of z calculated using the cross section, that one obtains tan a = tan1-53.1°2 -3the cross 4 same answer. above,the verify, using the geometry of section, that one obtains z 0.2667110 2 m E D the same answer. a = -79.4°2.25 Ans. D This result is shown in Fig. 6–33c. Using the value of z calculated above, verify, using the geometry of the cross section, that one obtains 0.20 the same answer. 4.95 4.95 B z 0.2-z 0.1-z C 2.25 B PAGE !88 C z

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