1 determi’le the slop e at point D Using the met hod of work and energy, decima l places_) 1 points eBook H int Print A __ __ __ __. D 2E I L ‘ 11 I I I L/ 2- — — –L /2 ,- –, .- The slop e at end B is I ~-~ MoL ~ . EI 1 caused by the couple Mo- (Round the fina l ans wer to thre e 2 g E = 72 GP a, um a llo y. Us in d urin g a ten sile tes t of an alu min ed ain obt a le nu mbe r:) dat from wn dra n has bee l ans we rs t o th e n e are st wh o fina the und (Ro y_ The stress-strain diag ram sho wn allo the of the mo dul us of tou ghn ess th e mod ulus of resi lien ce and dete rmine u (MPa) point s 600 eBoo k H int Print 300 150 I I I \ 0.006 I I I I I I I I I I I I I I 0.14 0.18 I I I I I I I I I I I The modulus of resilie nce of the ~- – ~ ] kJ/m3_ alloy is/,- The mod ulus of toug hne ss of the allo y is [ ] M J/ m3_ 3 Determin e the modulus of resilien ta) Stai nles s stee l AISI 302 tann eale d): £=1 90 GPa: poi nts ce for eac h of the follo win g met ‘ ay= 260 MPa I eSool< Hint Prim als: tb) Stai nles s stee l AISI 302 tcold -roll ed): E= 190 GPa l I, : ay= 520 MPa le) Mall eabl e cast iron : E= 165 GPa : ay= ~O MPa (Round the final answ ers to one dec ima l plac e.) The mod ulus of resil ienc e for AISI 302 tann eale d) stee l is\ The mod ulus of resil ienc e for AISI 302 tcol d-ro lled ) stee l is\ The mod ulus of resil ienc e for mal leab le cas t iron is\ \ kJt m 3 . \ kJ/ m3. \ kJ/m3_ 4 1 points eBook Determ ine the modulu s of resilien ce for each of the followi ng alloys: 6 (a) Titanium: E= 16.5 x 10 psi: ay= 120 ksi 6 (b) Magne sium: E= 6.5 x 10 psi: ay= 29 ksi 6 (c) Coppe r nickel (annealed): E= 20 x 10 psi: ay= 16 ksi (Round the final answers to one decimal pla.ce1.) Hint Print 3 The modulu s of resilience for titanium is ~I- – ~ llin.-lb! in . The modulu s of resilience for magnesium is I~–~I in.-lb/in3 . The modulus of resilience for copper nickel (annealed) is ~I–~I in.-lb/in3. 5 of the = 6 m/s whe n it strik es a sma ll plate attac hed to end A AB and has a spee d rod rm unifo the along s move O collar g A 3.5-k d iame te r that can be use d rod is 250 MPa , dete rmin e the sma llest the in s stres able allow the that ing know rod. Using E = 200 GPa and al place .) for the rod. (Round the final answ er t o one decim vo points Vo A eBook H ,nt P rint G I I it 7 D 1.2 m ·1 I The small est diam eter that can be used for the rod is I I .__—‘ mm . 6 points eBook Hint Prim rgy of the pris mat al stres se s, dete rmin e the stra in ene Taking into acco unt only the effec t of norm es_) plac answ e r t o thre e deci mal shown.Given: wo = 6w . (Round the final ~} 1111 r1 tJ t B ~ – L —-i►I h The strain ener gy for the b eam is re- –]- ~f ic bea m AB for the loa din g 7 show n. Take P= 7.6 kips_ (For S8 x due to bend ing for the steel beam and load ing Using E= 29 >< 106 psi. deter mine the strain energ y whole num ber.) 4 stresses.) (Roun d the final answ er to the near est ~8-4, I= 57.5 in .)(Neglect the effec t of shear ing p palms eBook Hint Print ” . [,,, L • J X D •·’ L (if1 _ \_3l1 S’S X \ !,;A .-\b. The strain energ y due to bend ing is \~_ _ _.\ in 8 points n ing th e ex pr es sio ifo rm thi ck ne ss _ Us un nno of on cti of a se th th e le ng irc ula r cro ss er AB , wh e re ds is mb er AB ha s a no n-c me mb al me ric of ind cyl ist w tw llo of ho fin al a n s we r ne the an gle The thin-walled cti on _ (R ou nd th e erg y de ns ity, de ter mi se -en s ain os cr str ll the wa for e th on e of ~ and the ex pre ssi T se d by the ce nte r lin 2 an d A is the are a en clo on cti se ss cro ll wa the , small element of .) to three decimal places = T’ eBook Hin t Prin t ~ ~/’ < els , ].~✓ \~ ~ Th p .:inn Ip nf twi ,t of th – – -7 J l J,’ ~ rP mP rn hP r AR is 1 X 8 :, ::.e LLIU I I UI I IUI I-LI i LU Id I LI u::,: 11e1 11ue 1 MD I Id::. d \I I t:::.: :,. u::,1 1 1y LI It: e.11.µ1 e::,: :,1u 1 I the len gth of a of me mb er AB, wh ere ds is st twi of gle an the e min ter ain-energy density, de the fin al an sw er and the expression for the str wa ll cro ss se ctio n. (Round the of line r nte ce the by enclosed ss section and A is the area small element of the wall cro to three decimal places.) , , ,e ll 1111 -VVd llt:U I IUII UVV Lym IUI ILd l I ,- . 2;A T’ 1 I IUI 1- UI IIIU I 111 LI IILI poin ts eBo ok H int Prin t X The ang le of twi st of the me mb er AB is I T I TL2 :r,t tds . ‘-·- – – ‘ GA 9 of th e bra ss owing tha t po rti on CD Kn a). GP 39 = G ( 8D to the bra ss rod TA = 76 0 N -m. = 26 GPa) is bo nd ed of the tw o rod s. Ta ke y erg en The aluminum rod AB (G ain str al tot 40 mm. de ter mi ne the has an inn er dia me ter of points eBook Hi-l t Prin t 37 5 mm / A 40 0m m / Th e tot al str ain e ne rgy of the tw o rod s is lr’—-_ -~ ___,] J _ ro d is ho llo w an d 9 s rod po rtio n CD of the b ras 39 GP a). Kn ow ing tha t = (G BD rod ss bra the N-m . 26 GPa) is bo nd ed to o rod s. Ta ke TA = 76 0 The aluminum rod AB (G = al str ain en erg y of the tw tot the ine erm det , mm 40 has an inn er dia me te r of poin ts eBo ok Hint Prin t 375 mm / A 40 0m m / The total stra in e ne rgy of t he tw . o rod s 1s [___ J ___, J . is ho llo w an d 10 ing E = t ruc tura l ste eL Us t o f a sp eci me n of s tes sile ten a g rin du dat a ob tain ed tw o de c i m a l w n has bee n dra wn fro m h e fin a l an sw ers to sho t m nd gra ou dia (R L train tee ss-s s e stre h t The e s s of the mo dul us of tou gh n 6 mo dul us of res ilie nce and 29 x 10 psi, det erm ine the places.) points
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Chapter 6 1. What are the primary differences between discrete and analog sensors? 2. What is the basic principle of operation for an inductive proximity sensor? 3. What is the basic principle of operation for a capacitive proximity sensor? 4. Compare the maximum ranges for proximity and photoelectric sensors. 5. What are the three primary sensing modes available in photoelectric sensors? Chapter 7 1. What is the difference between hydraulic and pneumatic actuators? 2. Explain the operation of a resistive temperature detector. 3. Explain the operation and construction of a thermocouple. 4. What is the primary advantage of a pyrometer over other temperature sensors? 5. Explain the operation of displacement flow sensors. 6. Explain the operation of the thermal mass flow sensor. Chapter 9 1. How do you control the magnetic field strength of an electromagnet? 2. When is counter EMF produced in a motor? 3. How is rotational direction of a DC series motor controlled? 4. How is pulse width modulation used to control the speed of a motor? 5. Contrast dynamic braking, regenerative braking, and plugging as ways of stopping a DC motor. Chapter 10 1. What is AC motor slip? 2. Why would an AC motor have a centrifugal start switch? 3. Why is the field generated by a three-phase motor stronger than the field produced by a single phase motor? Chapter 11 1. Describe the operation of a Hall effect sensor.
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Viscosity Notes From the Physics Hypertextbook https://physics.info/viscosity/ Discussion definitions Informally, viscosity is the quantity that describes a fluid’s resistance to flow. Fluids resist the relative motion of immersed objects through them as well as to the motion of layers with differing velocities within them. Formally, viscosity (represented by the symbol η “eta”) is the ratio of the shearing stress (ƒ/A) to the velocity gradient (Δvx/Δz or dvx/dz) in a fluid. ⎛F⎞ ⎛Δvx⎞ η =⎝ ⎠÷⎝ ⎠ A Δz ⎛ ƒ ⎞ ⎛dvx⎞ η =⎝ ⎠÷⎝ ⎠ A dz or The more usual form of this relationship, called Newton’s equation, states that the resulting shear of a fluid is directly proportional to the force applied and inversely proportional to its viscosity. The similarity to Newton’s second law of motion (ƒ = ma) should be apparent. F Δvx =η A Δz or ⇕ F=m ƒ dvx =η A dz ⇕ Δv Δt or ƒ =m dv dt The SI unit of