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T Interna) Forces and Moments (0) Iy prerjous Chapter, we Scuv how to detesmine the forces exeperienced by various members of a spructtue , frame en & machine . Now, in thts chapten, we wht see that how the loads apptied at venioul peints ©n A member determine the loading acting within the member TE is necessary fon w to cleterrs} ne the interna) foads jre Interna) ferce? LB weseremee acting inside the rmembey because, when designing Such a merrbes| we want to ensure that the materia vesists/ wohthsteunde = tnis lecclng- & To determine interna) teading , we use the meted ef section’ » hes We divide /eut the member at a peint where we want to find the forces &, tren apply ea.* g chapken , we focus of equi« mainly pin th 1 interna) loadings developed in ol peamy & cables ' consider fetlouing beam! wm consider Fellow an H " ain A 7 B NY oO section through B? bet us say that we want te obtcin the jnterna tocctings acting on the crocs section ak poiné B+ Then, we wf pass fr P, Ne ma 2 8 Vp Va = shean ferce —t to bajance forces Pn vertical dheotion + Ng = nexsmo) force —r te balance ferces In honzonte] Aires Mp = moment —> te boyance moments due to eterna) loads: From Newton's air Lows 8 Ax "3 ams Ns Ma A * We can we any portion of the beam to find eut the intesna) fead- ings ar @ pesticufcun pointt - * Fora straght & forte memben, the fFereet aot aeng the cus 6f- @ the member & ave equa} In magnitude Know - ‘ad force member only expert as we ences a nese) fesce ( Tensite/ comp’): So, Interroy foodng is ently a norma] force + # But, a2 force memben which fs nek straight wit! etpenience cll B typo + \ teaclings Inside tt 8 F e Sign Convention: nae) ) The force component that acts we” to the crog- section is catled 3) “The couple moment” / bending rrement 1) a) ro —7 Hh ie fe te ber tne segment concart upward . by M
NXO" 3X) The ferce component whieh actt pasoue) to the crosy— section }s cated os Shecun Foree (¥) - a) FH shear Ferce tet te rotate the member clockwise —et Vro hy Tt the shean ferce tre to yotale : Qt: Determine the nosma} fore, a Fence, and bending moment acting | Juse to the left, e point B » and use £0 the right , point c, of the 6 kN ferce ON the beam- | 6kN | 1 be 3m +h} 6m ——9} — Solution | ows Vx Bending moment~ wie 8 "Mn B Ny <> Noma) Force . Aan _ d Bt Wt F Fuing = 0: Ve > Shean fFesce- _ . ‘\ A FF (~ 6-875 K) Ft (H 18-5) 20 Vo —? Shean force, = . n CORa s (3:5) fe 3765K 8-3- The unifeorn sign Shown below |, = Mat ™ + eM has a mass of 65e kg & ie supported Woboue A Fulrg ebour 4 = 9, ona Fixed column . Design codes indicate that the expected mardmum vig IN LN IN Mw r J k 5226 ° 3 S&S | | | | | a mw Q Weaa-= to 4 k : © 5 535 H95 © © n Ww + Mpuing © | (6 3x6-3765°) shear & nosmal forces » iA + & (13-85% 3) -J (13-5 x5-25) A es My + pudnd £719. (295 j aA Al; = TEBIBL + 405K iN ‘ a4 Ma © 1901295 ) + 70875 405k Respective Cormpenents con be teamed a bending , twisting moments 4 cu Q:4. Fon the frame and loading ghown, determine the Interne) force 60 — q Solution : reer Ne Reaction exested by goound en reller. Ay, Ax—? Reactions exerted by leave Attached to the pin at A” ai Applying equi. conditions - zsMpA FO: 2 [Rea], J Ne XK OF1€ Ax ¢2502 250 => | Ax 50 # Now feb us fook at the Peps of pins and voller ®3 > FBO oF pinar c) CanHleven Beam. ——— 9 Staticatty Indetesmmate Beams * { | i | | | AS. Bessel | r hy hy 7 ’ \ \- 1 ®) Continuous Beam. —— >) Beam fixed at one end & simply Supported at the other end, '|# The distance *L’, between supporte is, ‘lent to detesrnine the reactions » we L °c) Fixed Beam- —— called the Spans the reactions ase dettominate tf the Support Involve only 3B unknowns Fe more unknowns are involved » the race trons are statically indeterminate , & the methed of Stetice Are noe SufHC- (This ts solved in Som) - # Beams supported by only two votlers ax posticly constrolned & more unden cestaln types ef leading: » # Sometimet 2 OF more beams cure connected by hinge to form a single continuow structtut- took at fol & Such structurt, there are > rer unknownt & thege 4 unknowns camnet be detemined by using the FBD of the whole Shucture, Howerer, ewe can dettr Mine the reactions by considering the Free ~ body Hagrar of each beam s Cpouctkety | Ag bron _! *® Sheam Force dH Bering Moment Diagrams (sep & Brip)- We saw eanlien that we can find interna) Forces /Jeading for a bear by using method of sectioning the beam at the deste tecakion & apply | in eq? of equjtioneumn P ' ' Ax by | ri Pia t ' 1 T i \g @= y tix} p/e 50,08 shown in abere diagram, we can find shean forte & bending mem ent: acting theoughouk the beam by Considering diagrams Like there. we can actually \ explained eardien during this proceit + he, ee con ea more dor Nght to fest op lett to right & show vaviation yn shean force & bending Mmomenk. wstth > # The graphs which shew tne -varfation ot shean force & bending moment with the distance measutd from one eng ef the beam }e calted ar Shean Force plagrom (SF®) & Bending Merent: Dioprom COM) vespecthety: #* We can form segment wise equations fon SF & BM ww a Function of distance: FaS are fosmred Segmentwice because SP & BM cepend on extesnoy lsoding conditions » we cht follow the sign convention * SPD & BMD fom above cone ! ' menduat dX orn elt ov ight e | Pla Ae Ha Reis serie 4 7 aS ; “Pls | Br A ee \ | PHA | ] S . rt 1 | | | SFo ait Osx <8 Bm. = KAR a o SH MG 3 8 HB) gh dx q(SF) 2 2m” dn # Relation b/w Load, Shean & Bendi- fe) Moment” Considen the ‘Fottowing beam’ LLB s oi A ID The beam is leaded with a nena uniforre leading having o magnitude of w andes / uni length . Now , considen the postion cD of the beam situated at Ask X from & w Vie a Cray yr AV | nN We assume that this ie how leads wy AK are acting ©n the beam. Al} art wre as per oun sign convention . Now, this CD part of the beam ut] be Tn equi: a® the beom fe Tn equi. Ye w DK + ht Av DV 5 — OOK een) On ow 4x20 pre. Lim SY HAV, an n AxtO uw = Intensity ot Slope of the shean force diagram is “ve of the dis fribuked food magnitude] ' “a acting en it: CPotensity of aishibuted lead’) : “ dv = —wdy + ve Ko {av é frweo dx “ x Re 4vVeN\-Vy = Jreede dee ad change In Asta unden | shean a sedieg c Change in shean b/w & pernts ts equ~ alto the arta under teading Curt b/w hose 2 pejntt oO * Above 2 eq ave only applicable ‘to the case where the looting on beam ts dis(ributed fecdting, when concent ~ Yated lead) force actt on the beam, then we cannot aire | apply these _ weokty ply | RV Now, considen moment equilbsium: —— —— firme = ww Axe AX 4 of ONY 2 Ams ViAK = wax)” cs OAM Lye WrAK Ax - ; dim as a =V° Axe Bx dr 4a =. qx Cc Slope of the bending moment diagram | hs equa) to the shean force. & dre Vqx Ms Ra Too = fecw-as ™) & rn . [Am = mm = f Cx) | x change in ec ara unde shean a Above 2 eg$ oF shean force & | eq! ef bending moments ast appti- cable to the beam only when dishi— buted food js acting en ad When a concentrated Porce or O couple moment ackt @b a print, it crtate: discontinuer Pe. jumpt in the sheam & moment diagrams & hence we how to teak FE Sepovalely, # Note thar while areiving ar the AY 2 ww ey, we assumed ew acts dx downward. Bur, if mctucilly re ack . upward Enen we with pur —w > 4 eqrutt| become IY) he us % Considen a concentoted force a. moment Casey - Fr Me L ng PO | = Ptypev = fove-F] vs due to Concentreuted feree the shea Feree diagroum wht sudenty chounge yee. A Jump . 50,)f we ave roving froro LR, then F acting Fownweere , wh make moment diagvam- shear fwrce diagram) reuue furp dewn ® ka C&S SF =X 15K (x measured freme) dv nets 7.5 — EME | an SF s—2 =]-5( %2) bos A,C , noon ts poy5% — (x meat: from 4’) By = ee & 4M) 2 ae Ax va dx = eaxderF> RN om M.-MA = oso , Ors ble BC Bm. HOH (1S) CRA) CM) mz fn - 15 HEA RAR ron OTT HZ TK By = — 075K EHS ACBM) ~ 15m rH) (i 5 measured TOK | fro A) | =v (bh) BC). os COM) 4 = Ag+ By = [tool =Mpr0 ————= = Boog [2004 600 xg = By x4 3800 —_r S50N, “Ry é NAYS FON . j a © ® Conclusions about SFD & BMD - ) The intesna) sheam & bending moment Functions & hence the diagrams, or then Slopes witt be discontinue, at | point where a dishibuted. Toad change on whert concentrated forces om couple momentt ast oppuied. a) When concentrared force acts: TE bring jemp dds eontinel ty TO. SPD - b) Concentrated force changes the Slope ef the moment clagram- ©) Concentrated moment wont affect SFP + dy Concentrated moment wht inproduce _fump aie continuity jn BMD: “3 When a beam ts subjected te conce- “ptrated loads ely , the shean is of constant value between Joads & bending moment vosiel Linearty petween foade » 3) Fox dishtbuted treading condition, Ww Hw » BV ST PeOOdK. dx ! : am. ome [ven de. dx 4) Physically , lOAds are Ofways app died oven a finite arta, & if the acttal too variation could be accountel fon, the sheam d moment diagrams weuld then be continuew oven the shatt?s entire Jeng th » 5) SFP & BMD ave very Useful: For designing as they help us to calidad fhe veuiation of interna) leading a well a, we can we ther find max- shean force >» bending moment & thein location + 6) The eqs of shear & bending moment | Currel are ayfwoys, respectively , one 4 two degrees highen than the eq. of tne [oad curve. 7%) where bending moment 7s mastimum, AM av eo . dx Tee, Tap , AY, Ax, u By, Bx, Yo, Ip 9 unknowns . & ak each point BB -CrD , we can write 9 eq? ef force equilibrium - - 80, In total, we have B eq! ef equi. | Bue to complere the Solution, we need td mor eqs fon eg, we can get thit one mort eq. by utilizing the known lengtr L op cable & hence we have 9 eq! fer Q unknowns - Cour we this eq: complicates things & hence most ef the times, “ene of the Y waluet (eg Yer 3) it mentione) (ie: 8 unknowns) « | Following question will make ereryt~ hing cjear + =< Solution ‘ Ose First, consider tne whole Sfructurt ZFys0 => Ayr FY = 22 KN GQ =Mp 20 SMA =O ARSHIBXB A TKI TEX S Eys t+ ott ® oan, 18 Ay = ARN, Ey lOkN now, considen Following Subpoot of the cabjes - =Me 50 Ax dt Axx i = thx B Te] AK 6.33 — ® KN | (ii Speier NE i ee Now, wee Top x Cos Ocp = 6.35 = Ape: iz Toe Tep SinGep = [Seorie= T+ ® | Toe ln Spe = 12 — [ore = 8) = BY.66 tanOep = —/ Oed = AT-88°| | We 083 ODE = 6-92 G=38 toe = 12835 KN | Tee = 9-438 kN a A very important poine to nett “ » FRD- = wren nee ts thet , hovizonto] component: Ay 2 JZKN 1 633 On of tension in each cable if equal §P SP £0 rr) Ax Cor Ex)» why ¢ ae Pap. cosOnn = 6.35 _© Pap =TAe a To venfy this, consider FBD oF the port of the Shuckurt Ce done to =e fe = 63-19). i ‘ . tanS ap = 18957 re a tig. ©: Fap S)00aR = I + @ i] Fae o-Tas = 13-57KN e= angle measured with ore oe : ental. tonkae oe = fze = 5-800 | cis St hs slope (telaked to slope) + | 8 ; ep OTe Ax 180 gap Ter. Tac SinS ac =15 ows eep _ E Tee cor8pc = S:S* <7. T witt be max when © Ts max, SKN toate ee v. Tmax = Ax Coo Omar, — Jove FS kN Tec a ie: | .%-] Trase = [3-567 KN | % = Tap - x a er f ( Seem): xe) fax $ — ® This eq: is used to cetermyne the curve fer the cable, he ys FO) The herizenta, Force component Fr and the additional 2 constant , pS q ECs y vesulkiing frorn the Tntex / gration are determined by applying, | the boundouy conditions «fer ‘the curve- | mw N-2- =—_ we 0) % measured in the x-dincten C= lewem mest point~ ' Sfero te Tees®=To —— sFyeeo Fe TSINO = pesubkane 16 diners force due to ditfibuted lead, (on the post of sting’): shown + w & tane = _W_ A Te ° . Toos@ = To 7 TSINO = ws T= Jrtrws , tone = To tone = Ay = Slope ar any peint en ax Coub]e. ay. ao = he ~— Same as eq, ceve}oped im m-) 2 , au the ey! ave exactly tne same. # Tf toading is uniform then, x We { ws) dx s o wom + & erigin he ak the lowest point of : the cable. = wr * We wl) Xk" T= [Tne Curve Formed by cable toaded uniformly aleng the herizonty he a faraboja Cabfes hanging under then own a male * | TCS xe welght are net foaded uniformty alory the horizonta] & do not fom powrabolag Howewen, the erron Introduced by Assuming & parabotic shape Fon cables y hanging under theln own we} gh is smal) when the cable js Suftic) ently tout” ® Length of Coble . (dsyt= canyr (dys os = NCA (aah é | *. 8 = Length ef cable ‘= fds ds = It dy on x! ole . ( ) Length ef cable % Umit wit depend on cable Segment & origin» ae 
