807 - Fundamentals Of Engineering Exam
I The Fundamentals of Engineering (FE) exam is given semiannually b} the National Council of Engineering Examiners (NCEE). and is one 01 the requirements for obtaining a Professional Engineering License. A portion of this exam contains problems in mechanics of materials. and this appendi.'t provides a review of the subject matter most often asked on this exam. Before solving any of the problems. you should review the sections indicated in each chapter in order to becomefamiliar with the boldfaced definitions and the procedures used to solve various types of problems. Also, review the example problems in thesesections.The following problems are arranged in the same sequenceas the topics in each chapter. Partial solutions to all the problems are given at the back of this appendi.'t. Reference: .Mechanics of Materials. by R. C. Hibbeler, 3rdEdition
AP. D REVIEW FOR THE FUNDAMENTALS OF ENGINEERING EXAMINATION Ch.pter l-Re\1e" All Sct.-tiun!i 0-1 Detenninethe resultantinternalmomentin the memo ber of the frame at point F. 807 D-6 The ban of the truss each have a cross-sectional area of 2 in.2 Determine the average normal stress in member CB. 0-7 The frame supportsthe loadingshown.The pin at A hasa diameterof 0.25in. If it is subjectedto doubleshear. detenninethe averageshearstressin the pin. 0.75ft Prob. 0-1 0-2 The beam is supported by a pin at A and a link BC. Determine the resultant internal shear in the beam at point D. 0-3 The beamis supportedby a pin at A and a link BC. Determinethe averageshearstressin the pin at B if it has a diameterof 20 mm and is in doubleshear. D-4 The beamis supponedby a pin at A and a link BC. Determinethe averageshearstressin the pin at A if it has a diameterof 20 mm and is in singleshear. 0-5 How many independentstresscomponentsare there in three dimensions? Prl/b. 0-7 D-8 The uniform beam is supported by two rods AS and CD that have cross-sectionalareas of 10 mm2 and 15 mm2. respectively. Determine the intensity w of the distributed load so that the average normal stress in each rod does not exceed 300 kPa.
808 AP. D REVIEW FOR THE FUNDAMENTALS OF ENGINEERING 0-9 The bolt is used to support the load of 3 kip. its diameter d to the nearest t in. The allowable normal stress for the bolt is UalkJw - 24 ksi. Determine 3 kip E.XAMINATION Chapter 2-Review All Sections D-13 A rub r band hasan unstretchedlength of 9 in. If it is stretchedarounda pol havinga diameterof 3 in. determineth averagenormalstrain in the band. 0-14 The rigid rod is supported by a pin at A and wires BC and DE. (f th maximum allowable nonnal strain in each wire is E;lJ w 0.003. detennin the maximum vertical displacement of the load P. Prob. 0-9 D-I0 The two rods support the vertical force of P 30 kN. Determinethe diameterof rod AD if the allowable tensile stress for the material is Uallow 150MPa. 0-11 The rods AS and AC have diameters of 15 mm and 12 mm, respectively. Determine the largest vertical force P that can be applied. The allowable tensile stress for the rods is UaUow ISO MP:l. D-15 The load P causesa normal strain of 0.0045 inJin. in cable AB. Determine the angle of rotation of the rigid beam due to the loading if the beam is originally horizontal beforc: it is loaded. 0-12 The allowable bearing stress for the material under the supports A and B is UalkJW 500 psi. Determine the max- imum unifonn distributed load w that can be applied to the beam. The bearing plates at A and B have square cross sections of 3 in. x 3 in. and 2 in. x 2 in. respectively. "-6ft '!';.'i"" t.-" Pruh. 0-12
AP. D REVIEW FOR THE FUNDAMENTALS 0-16 The squarepiece of materia!is deformedinto the dashedposition.Determinethe shearstrain at comer C. OF ENGINEERING EXAMINATION 809 D-23 A l00-mm long rod has a diameter of 15 mm. If an axial tensile load of 100 kN is applied. determine its change in length. E 200 GPa. 0-24 A bar has a length of 8 in. and cross-sectional area of 12 in2. Oetennine the modulus of elasticity of the material if it is subjected to an axial tensile load of 10 kip and stretches 0.003 in. The material has linear-elastic behavior. 0-25 A 100mm-diameter brassrod hasa modulusof elasticity of E - 100 GPa. If it is 4 m long and subjected to an axial tensileload of 6 kN, determineits elongation. 0-26 A lOO-mmlong rod hasa diameterof 15 mm. If an axial tensile load of 10 kN is applied to it, determineits change in diameter. E 70 GPa, v 0.35. Chapter 4-Review Sections 4.1-4.6 0-27 What is Saint-Venant'sprinciple? D-28 What are the tWo conditions for which the principle of superposition is valid? Chapter 3-Revie,,' Sections 3.1-3.7 0-17 Define homogeneous material. 0-18 Indicate the points on the stress-strain diagram which represent the proportional limit and the ultimate stress. D-19 Detenninethe displacement of endA with respectto end C of the shaft. The cross-sectional area is 0.5 in2 and E 29(103) ksi. 0' D A.J--.d' ""' -- oO 2 kip 6 kip 4 kip ,\ B C \ A 2ft '! B6ft Prob.0-29 Prob. D-18 0-19 Define the modulusof elasticityE. D-20 At room temperature,mild steelis a ductile material. True or false. 0-30 Determine the displacement of end A with respect to C of the shaft. The diameters of each segment are indicated in the figure. E 200 GPa. D-2l Engineering stressand strain are calculated using the actual cross-sectionalarea and length of the specimen.True or false. 0-22. If a rod is subjected to an axial load. there is only strain in the material in the direction of the load. True or false. Prl.h. 1)-.'0
810 AP. D REVIEW FOR THE FUNDAMENTALS OF ENGINEERING 0-31 Oeterminc: the angle of tilt of the rigid beam when it is subjected to the load of 5 kip. Before the load is applied the beam is horizontal. Each rod has a diameter of 0.5 in. and E 29( Io!) ksi. EXAMINATION 0-34 The column is constructed from concrete and six steel reinforcing rods. If it is subjected to an axial force of 20 kip, determine the force supported by the concrete. Each rod has a diameter 4.20(103) of 0.75 in. Ecorr ksi. Esr 29(103) ksi. 6 in 2ft 0-32 The unifonn bar is subjected to the load of 6 kip. Detennine the horizontal reactions at the supports A and B. :.: D-33 Thecylinderis madefrom steelandhasan aluminum core.If its endsare subjectedto the axial force of 300kN, determinethe averagenormalstressin the steel.The cylinder hasan outer diameterof 100mm and an inner diameter of 80 mm. Est 200GPa,Eat 73.1GPa. .Prob.0-34 0-35 Two bars. each made of a different material. are connected and placed betWeentwo walls when the temperature is Tl - lSoC.Determinethe forceexertedon the (rigid) supports when the temperature becomes T2 25 C. The material properties and cross-sectionalarea of each bar are given in the figure. Brass E.,- IOOGPa Steel Ell . 200 GPa a.,' 12(lo-6)rC A., 175mm: Proh. 0-33 A., 300mm: c B A ,,' a., - 11( loo.6)fOC Inn 1\N mm .or Pn)b. 0-35 : 2OOmm---i
811 D-36 The aluminum rod has a diameter of 0.5 in. and is lttached to the rigid supports at A and B when TI - gooF. If the temperature becomes T 100 F.and an axial force ){ D-4O The solid 1.5-in.-diameter shaft is used to transmit the torques shown. Determine the shear stressdeveloped in the shaftat point B. p s 1200Ib is appliedto the rigid collar asshown.deter- mine the reactions at A and B. ae/ 12.8(10-b)rF. Eel 10.6(1Q6) psi. frob. 0-40 frob. 0-36 0-41 The solid shaft is used to transmit the torques shown. Determine the absolute maximum shear stressdeveloped in the shaft. Prob. 0-37 Chapter 5-Re,ie'" Sections5.1-5.5 0-38 Canthe torsionformula,l' TclJ,be usedif the cross sectionis noncircular? 0-39 The solid O.75-in.-diametershaft is used to transmit the torques shown. Determine the absolute maximum shear stress developed in the shaft. frllb. 0-:\9
812 AP. D REVIEW FOR THE FUNDAMENTALS OF ENG(NEER(NG EXAMINATION 0-43 Dl:tl:rmin\: thl: angll: of twist of thl: l-in.-diameter shaft at end A wh\:n it is subjl:cted to thl: torsional loading shown. G 11(10-')ksi. 0-47 Thc: shufl is mudt: from u stt:el tube having a b core. If it is fixcd lO tht: rigid support. determine the al of twist thut occurs ul its c:nd. Cst 75 GPa and G 37 GP3. ( Pruh. D 3 I'rl,h. O-J7 D-44 The shaft consistsof a solid section AS with a diameter of 30 mm. and a tube B D with an inner diameter of 25 mm and outer diameter of 50 mm. Determine the angle of twist at its end A when it is subjected to the torsional loading shown. G 75 GPa. D-48 Determinethe absolutemaximumshearstressin shaft.JG is constant. Do.i, :- Prl.h. Prllb. D 0-45 A motor delivers 200 hp to a steel shaft. which is tubular and has an outer diameter of 1.75 in. If it is rotating at 150 rad/s. determine its largest inner diameter to the nearest in. if the allowable shear stress for the material is "allow 20 ksi. D-46 A motor delivers 300 hp to a steel shaft. which is tubular and has an outer diameter of 2.5 in. and an inner diameter of 2 in. Determine the smallest angular velocity at which it can rotate if the allowable shear stress for the material is T"uuw 20 kosi. Chapter 6-Review Sections 6.1-6.5 0-49 Detennine the internal moment in the beam a functionof .\',where2 m S .\' 3 m. kN/m "., 2m : Prclh. O 9 1m Im--:
AP. D REVIEW FOR THE FUNDAMENTALS 0-50 Determine the internal moment in the beamas a functionof x. 'here0 s x s 3 m. 0-54 OF ENGINEERING 813 EXAMJII;ATION Determinc the maximum moment in the beam .x.N ! 4m I -"- m I ;2.--; Prob. D-5.a D-SI Determinethe maximummomentin the beam D-S5 Determinethe absolutemaximumbendingstressi the beam. 8 kip 8 kip D-S2 Determinethe absolutemaximumbendingstressin the beam. 1-4ft I -2Olin. -. in. -I14fi--l 6 ft Prob. 0-55 0-56 Determine the maximum bendingstressin the SO-mmdiameter rod at C. 16kN 400 N/m Prob. D-52 D-S3 Determinethe maximummomentin the beam. A. .L.B L ," - m . I -, I Prob. D-56 Prnb. 0-53 0-57 What is the strain in a beam at the neutral axis?
AP. D ( REVIEW FOR THE FUNDAMENTALS OF ENGINEERING D-S8 Determinethe momentM that shouldbe appliedto the beamin order to createa compressive stressat point D of to ksi. 0-61 EXAMINATION Determine the maximum stress in the beam's cross s ction. -"!" in. .1. .,--- Chupter 7-Revie" St.'cticm 7.1-7.& 0-62 Determinethe maximumshearstressin the beam. Pnlh. I)- Determine the ma. imumbending stre in the beam. 11 V.20tN' 200 mm ! i t-- :,A' l50mm" I Proh. 0-62 3 The beamhasa rectangularcrosssectionand is sub 2 kN. Determinethe ma. imum shearstressin the beam. jected to a shear of V D-6O Determin the maximumload P that can be applied to th beamthat is madefrom a material havingan allow- able bending str ss of 0',,11ow 12 MPa. ' 150mm .1- p I '-- 0.5 in. .1 20 mm T 110.5 in. o.s ;0 mm -:-T 20 mm IOOmm I c ,:i .;;:r;i ii 2m ;,. PTllh. 1)-iu} -i-"'-- ::: '" m - -- 3. 1ft. 1'rllh. 1 -6.; . IjI.5 1ft.
AP. D REVIEW FOR THE FUNDAMENTALS OF ENGINEERING EXAMINATIO D-64 Delerminethe absolutemaximumshearstressin the shafthavinga diameterof mm. 815 0-67 The beam is made from four boards fastened to ether at the top and bottom with two rows of nails space:d ever)' 4 in. If an internal she:ar force of V 400 lb is applied to the boards. determine the shear force resisted by each nail. ( ! 3 . . -: 3 . . 11 Prob. D 0-65 Determinethe shearstressin the beamat point A whichis locatedat the top of the web. Chapter 8-Revie,,' All Sections D-68 A cylindricaltank is subjectedto an internalpressure of 80 psi. If the internal diameterof the tank is 30 in. and the wall thicknessis 0.3in. detenninethe maximumnonnal stressin the material. Prob. 0-65 A pressurized sphericaltank is to be madeof 0.2S-in.thick steel. If it is subjectedto an internal pressureof p s ISOpsi. determineits inner diameterif the maximumnormal stressis not to exceed10 ksi. 0-70 Determine the magnitude of the load P that will cause a maximum nonnal stressof Umax- 30 ksi in the link D-" The beamis madefrom two boardsfastenedtogether at the top and bottom with nails spacedevery 2 in. If an internal shearforce of V 150Ib is appliedto the boards, detenninethe shearforce resistedby eachnail. along section a-Q. I t-2 PrClIl. 1)-711 I :T -L . O.5m.
816 AP. D REVIEW FOR THE FUNDAMENTALS OF ENGINEERING 0-71 Determineth maximumnormal stressin the horizontal portion of th bracket.The bracket hasa thickness of 1 in. and a width of 0.75in. EXA "NATI0N 0-74 The the solidcomponents cylinder is subjected the loading shown. Oetennine of stressattopoint B. 0.75in. Prl)b. 0-71 0-72 Determinethe maximumload P that can be applied to the rod so that the normal stressin the rod does not exceed O'max Z 30 MPa. Prub. 0-7.& 20mm frob. 0-72 D- 73 The beam has a rectangular cross section and is subjected to the loading shown. Determine the components of stress O'x.O'yand T:t.V at point B. Chapter 9-Review Sections 9.1-9.3 0-75 Whenthe stateof stressat a point is representedby the principalstress,no shearstresswill act on the element. True or false. 0-76 The state of stressat a point is shown on the element. Determine the maximum principal stress. 6ksi 4ksi .5 . 17::-- In,1.5in. Prllh. 1)-7.; Pn)/). 0-76
AP. D REVIEW FOR THE FUNDAMENTALS OF ENGINEERING EXAMINAnON 0-77 The state of stressat a point is shown on the element. Determine the maximum in-plane shear stress. 817 D-8O The beam is subjected to the loading shown. Determine the principal stress at point C. IISOpsi -r7Smm , . -r- 100psi 7Smmi -1- 200 psi Prob. D-77 Pr(tb. D-80 0-78 The state of stressat a point is shown on the element. Determine the maximum in-plane shear stress. 130MPa Chapter 50MPa 12-Review Sections12.1-U.2. U.5 0-81 The beam is subjected to the loading shoWl!. Determinethe equationof the elasticcurve. 1 is constant. 2kip/ft f Prob. D-78 ) 0-79 The beam is subjected to the load at its end. Oetenninethe maximumprincipalstressat point B. B -3ft Prob. 0-81 D-82 The beam is subjected to the loading shown. Determine the equation of the elastic curve. 1 is constant. 3 -'!'--W--1.-:T '!I". ., ".-r--'-- A .'!;:ii,' 101\, Prclh. D-S2 c'.i! i
818 AP. D REVIEW FOR THE FUNDAMENTALS OF ENGINEERING D-83 Determin the displacem nt at point C of the beam shown. Us th method of superposition. EI is constant. EXAMINATION D-87 A 12-rt wooden rectangular column has the dimen. sions shown. Determine the critical load if the ends are as. sumed to be pin-connected. E - 1.6(103)ksi. Yielding does not occur. p ,." 1.1' 3ft 'I Pruh. 0-83 T 4in. 1 D-84 Determin the slopeat point A of the beamshown. Usethe methodof su rposition. 1 is constant. kN/m Prob. 0-87 D-88 A steel pipe is fixed-supported at its ends. If it is 5 m long and has an outer diameter of SOmm and a thickness of 10 mm. determine the maximum axial load P that it can carry without buckling. Est 200 GPa. O'y- 250 MFa. Chapter 13-Review Sections 13.1-13.3 D-85 Thecritical loadis the maximumaxialloadthat a column can supportwhen it is on the vergeof buckling.This loading representsa caseof neutral equilibrium. True or false. D-86 A 50-in.-long rod is made from a 1-in.-diameter steel rod. Determine the critical buckling load if the ends are fixed supported. E 29(103) ksi, O'y 36 ksi. - 0-89 A steel pipe is pin-supported at its ends. If it is 6 ft long and has an outer diameter of 2 in. detennine its smallest thicknessso that it can support an axial load of P 40 kip without buckling. En - 29(103)ksi. O'y- 36 ksL 0-90 Oetennine the smallest diameter of a solid 4O-in.-Iong steel rod, to the nearest tr;in., that will support an axial load of P - 3 kip without buckling. The ends are pin connected. ESt 29(103)ksi, O'y 36 ksi.
AP. D REVIEW FOR THE FUNDAMENTALS D-8 PARTIAL SOLlTTIONS AND ANS\I\TERS CD is a two-forcemember MemberAE: 0; FCD - 600 Ib D-9 w n' ) 1\1" W W 10' - -' 3 MN/m 0' - ;24 - -/4I; d - 0.3989m. use d - 0.5 in. - 0; A,. 6 kN SegmentAD: 1:F,. 0; V - 2 kN 0., T AS AM. D-1. BC is a two-forcemember. BeamAB: 1:Ms Rod AB: P us -' 300( A' Rod CD: 0; A , 800 Ib SegmentACF: IMF 0; MF 600 lb. ft 819 Beam: I.F.v IME EXAMINATION I.MA-o.,TcD 2w 0-1 Entire frame: IMs OF ENGINEERING Ans. Ans. D-4 SC is a two-forcemember BeamAS: - Ans. - 1:MA 0; Tsc 4 kN 1:Fx 0; Ax 3.464 kN - - 0; A" 6 kN FA - V(3.464)"l (6)"l- 6.928kN 'fA - fA - 6.928 - 22.1kPa A f(O:m)'i 1:F" D-S 6: O'z,0'7' O' ,TzyoTy:, T:z Ans. Ans. Ails. 1.8 w w -'(2)(2)' D-6 Joint C: :t.I.Fx - 0; Tc. - 10kip Ta u A 10 2 . Sksi 11(3)- 9 Ans. 9 0-14 - 0.0472in.lin. - .4ns. (8DE)max masIDE 0.003(3) 0.009m By proportionfrom A, 8sc 0.009 (t) 0.0036m (8sc)max - EmasIsc 0.003(1)- 0.003m 0.0036m Use 8sc - 0.003m. By proportion from A, 8p 0.003 0.00525 m - 5.25mm Ans. S - Ans. 1.11kip/it S - ( ) -
820 AP. D REVIEW FOR THE FUNDAMENTALS D-lS lAB \rf(4) (3) 5 ft /' ,IB S 5 5(0.(x 5) 5.0225 ft The angleBCA wasoriginally 8 OF ENGINEERING EXAMINATION D-27 Stressdistributionstl:nd to smoothout on sections - 9()0.Using the co- sine law. the new angle BCA (8') - 5.0225 '\1(3):' (4)Z 2(3)(4) cas 8 is further rl:movl:d from thl: load. 0-28 An.r. 1.) Linear- lastic material. 2.) No large deformations. An.r. 8 1,X).53Mo Thus 8' - 90.5)80 - 9()0 - 0.538 0.0166in. Ans. 0-18 Proportionallimit is A. Ultimate stressis D. Ans. Ans. 0-19 The initial slope o( the 0' - E diagram. Ans. Am. 0-10 True 0-2l False. Use the original cross-sectionalarea and length. AIlS. f 0.5(29(103» O.S(29(1f}J» -f 0-17 Material hasuniform propertiesthroughout. 4(6)(12) -2(2)(12) Am. Ans. 12(103)(0.5) (0.02)2 200(10'1) 27(loJ)(0.3) (0.05)2 200( 10'1) - 0.116mrn Ans. 0-31 BeamAS: M. 0: FSD 2 kip Fy 0: F"c 3 kip - - " !!::. 3(8)(12) . AE of(0.5)2 29(1OJ)- 0.0506In. s PL - Ai 2(3)(12) of(0.5)2 29( 103) - 0.01264in. 0-22 False.There is also strain in the perpendiculardirection due to the Poissoneffect. AIlS. Ans. Ans. (1(xx)()(8) F. 2 Ans. !.!S!:l 4 kip. FB 2 kip AE Ans. 0-33 Equilibrium: P l Pill - 300(loJ) Compatibility: 6(103)4 - P",L .tn [f (O.08)Z]73.1 (IQY) MPa Ans. 64.3 f (O.OI):!100(10"') - 3.06 mm - - D-32 Equilibrium: F", FB-6 Compatibility: F. (1) 6c'", 3aB: AE
AP. D .14 REVIEW FOR THE FUNDAMENTALS 821 D-J8 No, it is only valid for circular cross sections. Noncircular cross sections will warp. Ans. Equilibrium: PCQfIC Pst OF ENGINEERING EXAMINATION 20 Compatibility: - 40 Ib . ft D-.J9T- CD P (2) 8 '; ['IT (6)2 - 6(f)(O.7S)2] 4.20(103) -J - Pit (2) (103) - 6(f) (0.75)229 P- - 17.2kip Pit - 2.84kip Ans. D-.- ,.35 cSr-p Ia4TL 4 40( f 12)(0.375) (0.375) IC T- - 5.79ksi Equilibrium of segmentAS: TB 30Ib ft T I:;. 30(12)(0.75) 543 . B J f (0.75)" psi . - Ans. B.o8d I.fb. AE Compatibility: 0 12(10-6)(25- 15) (0.4) 21(10 )(25 - 15)(0.2) - (0.4) (0.2) 175(10-6)(200(109» - 300(10 )(100(109» 0 F 4.97kN Ans. )-36 Equilibrium: FA Fs 1200 Compatibility Removesupportat B. Require 8s (8B1A)telnp (8B1A)\oad 0 - D-4% - of200(12)(3)(12) (0.75)411(1 1') 0 of300{12)(2)(12) (0.75)411(1 1') - a4 TL "'" !.!::. 0 LAE 12.8(10-6)(100 - 80)(14) 1200(6) F.(14) f (0.5)2 10.6 (1C1') - f (0.5)2 10.6 (1C1') - - Ans. Am. a4TL P 1.86kip :.t ' I!:. . fG - 600(12)(3)(1 ) f (0.5)411(106) - f200(12)(2)(12) (0.5)411(106) f 100(12)(3)(12) (O.5t 11 (106) Am. -f 40(0.3) (0.OLS)4 75(10"') 0 f [(O.025 -20(0.2) (O.Oi25 J 75(10") .fb. 0 AE 12.8 (10-6) (SO- SO)(14) Ans. - 0.253 tad counterclockwise when viewed from A. Compatibility Removesupportat B. Require (8B1A)load - 0.0211fad clockwise when viewed tad 0 0-37 Equilibrium: FA Fs P SinceFs O.FA P 8s (8B1A)remp -0.0211 from A. D-43 Fs 1.05kip FA -1531b TL -400(12)(2)(12) 2: 1G - f (0.75)411(1 1') .AIB P(6) t (O.5:)21Q6 (106) - 0 Ans. - f [(0.025)4-30(0.3) (O.OI25).J 75(10") -0.209 (10-3) fad - 0.209(10-3) fad clockwise whenviewedfrom A. .4ns.
822 AP.D REVIEW FOR THE FUNDAMENTALS OF ENGINEERING EXAMINAll0N D-!G A" 3 kN Use sectionof length . P - 110(XXJ T- -;;; 1SO - 733.33tb . ft T.lkN rj dl j;Tc 20(103) - Intensity of w - t at . L 1:M -3 (t )[t.r)(t.r)] At. 0 f 733.33(0.875) [(0.875)4 - rj4] M 3.r-- 9 - 0.867in. - 1.73 in. use dj - 1.625in. It in. . ns. D-5l By Ay ", 2.6 kip - 4.6 kip Draw M-diagram Mmax- 7.80k. ft (at C) Ails T - .f. - 165 OJ D-S2 Draw M-diagram Mm. 20 kN . m (at C) ill.!!!! . (12)(1.25) - An - (1)4] T [(1.25)4 .-tns. (II a 54.7 rod/s D-47 Equilibrium: Tst Tbr 950 Compatibility: t - - Ay 2.33 kN By 6.617kN - Draw M-diagram Mmaxc 11 kN' m (at C) (0.6) f [(0.03)4 -Tst (0.015)4] 75 (109) T b, 0-53 - Tbr (0.6) f (0.015)4 37(1Q9) D-S4 Ay ,.\n,f By 800N DrawM-diagram Mmax 1600N . m (within CD) Tn 919.8N. m 30.25(0.6) f (0.015)4 37(109)- 0.00617 tad - Ans. D-SS Ay By 8 kip M- c 8 (4) - 32kip. ft I 0' D-48 Equilibrium: T,4 Tc-600 Compatibility: D-S6 T,4 (2) 881c 881,4;! .ill. JG - Mc T 32(12)(3) if (2)(6)3 - 32 D' An. J Ay z By 100 N Mmu 1250N m - . l?qvO.025\ , 5509MPa O'mu - -Mc I - .--, f (0.025)4 . T,4 200N.m Tc 400N.m Tc - 400(0.025) Tmax T f (0.02S - 16.3MPa A. 5.5kN Usesectionof length%. L !M 0; -5.5 % 4 (2)(% ,\{ 8 2.5% - - - 0-58 1) M .-\n.f An.\'. 0-57 . - 0 0-49 .\ns 30.25 N . m -0 \n\ 0' M M \n.\ - M(1) 10(103) [i'f (4)(4)3 -"i'I (3)(3 ] - 145.8kip .in. - 12.2kip' ft \11\
AP. D REVIEW FOR mE FUNDAMENTALS OF ENOINEERINO EXAMINATION -59 From bottomof crosssection - - 4«80)(20) 95(30)(100) - 7,S.870 Y 80(20) 30(100) DIm I ft (20)(80)' 2O(S»(7's.870- - 40)2 - -!i (100 )' 1 X9.5 - 75.m)2 4235(10-6) m - TMc - 0'- 10(103)(0.075870) 4 S'(10-o) . - 179MPa AIlS. - I. D-67 I-.l. 12 (6)(6}' - 34.66(10-6) m4 Mc (1)(6.5 . AM ,.- XR. /1 - 4(103)(8 84.667 (1)- 5») - 567pst F. q.r. (56.2SIb/in.) (2 in.) . 112.5Ib (0.02)(0.150 ' 2 [-& (0.1)(0.02)3 (0.1)(0.02)(0.(MS)2] 0" /- -k (1)(6)' 6(1)(5 - 3)' -k (8)(1)' 8(1) (6.5 - 5)Z- 84.667 in. (6)(4)3 . 32in. 12 15O(1)(6)(2)J - 56'tc Ib/iIn. q . .J 32 - PI2 (2) - P (at C) ft -Y - I.yA 3(6)(1) 6.5(1)(8) . 6(1) 1(8) - 5 In. D-66 J.l .a Ay- PI2 M- D-'5 From the bottom: M 12(Iv-) q - 1 P CO.(W5) - 34.66 (10-6) p - 4.38tN Ans. (0' )0 - ! 400(2.5)(6)(1) - qs - 34.62Ibfan.(4 in.) - 138lb t . ! (6)(4)J - 404 . pst (SOsin JOO)12 (2) - 69.23Ib/im. - 69.23/2 - 34.62 Ib/in. D-a 0'- l!. - (4) (6)J 86.67 Foronenail q F ).41 Maximum stress at DorA. (SO JOO)12 (3) - - .l.12 (4)(4)3- 86.67in4 0.3 - 4(XX) psi- 4 ksi Q is upperor lower half of crosssection. Tmu- - -lMPa v-a AIlS. 0-63 [ ,', (3)(4)3-,', (2)(3)3- 11.5in4 Q is upperor half of secUon. - (0.75)(1.5)(2) - 3.75 XR. [t - 11.5(1) - 0.UJ tsi. T- D-64 Ay - 4.5tN. By - 'rmu - It - 2(o:s Ans. -& (0.5)(2)3 Ans. P-3kip 0-71 At 8 sectionthrou8h the center of bracketon centroidal axis. N-700lb V-o 1.5tN section. M - 700(3 0.375) - 2362.5 Ib . in. t (0.06) . (0.03)2] t'r (O.03)C] 4.5(10')[( 2.12MPa M - (2 l)P - 3P P Mc 0'-- A I P (3P)(1) 30 ) in3 V- - 4.5tN (atA) Q is upperhalfof Ans. D-70 At the sectionthroulh centroidaJaxis N-P It Q - (1) (2) (3) Ans. 150, 2 (0.25) Ans. r - 33.3in. d - 66.7in. 2 Am. v- AP TMc .-tn.Oj - 26.1ksi 700 [1'r 2362.5(0.375) (1)(0.75)3J -0:7S'('i) Ans.
824 AP. D REVIEW FOR THE FUNDAMENTALS OF ENGINEERING EXAMINATION 0-75 True 0-76 0'. 4 ksi. u.y- -6 ksi.1'., -8 ksi Apply Eq. 9-5.0'1 8.43ksi. O'l -10.4 ksi 0-'77 0'% - 200psi.0', -ISOpsi.1'.,- 100psi Apply Eq. 9-7.1'max 202psi in.".a. 0-78 0'.- -SO MPa.0', - -30 MPa, 1'.,- 0 Use Eq. 9-7. 1'max 10MPa in-pla. 0-79 At the crosssectionthroughB: N 4 kN. V 2 kN. M 2 (2) 4 kN . m - - { .-tns. P-23.5N 0-73 At sectionthroughB: 500 lb. N V - 400 lb. M 400 (10) - 4(XX) lb. in. - 0'8-- --- Axial load: 0'. Shear load: 1'., . It I I 'TB . 4 - - 4/0.03) ., - , 0:03{0:()6) * 0.03)(0.06)3 224 kPa (T) - 0 sin :e Q - O. Thus 0'\ pst An.J. An.J. 224 kPa O D-8O Ay - By - 12 kN Bending moment: My 4000(1) 0'. Note 400(1.5)(3)(1)1.37.5. ,'I(3)(4)']3 Mc A - .f. A - 4 (3) - 41.667psi (T) - - P . Segment S ,::&(3)(4)1- 250pst (C) A 1I.r. A C: Vc.O. Mc 24 kN. m 'Tc - 0 (since V c 0) Thus - 41.667 - 2500', - 0 'Txy 37.5 psi O'x psi (q Ans. Ans. Ans. D-74 At sectionB: N; - SOO N. V, 400N. Mx 400 (0.1) 40 N. m M, 500(0.05)- 25 N . m. T; 30 N . m Axial load: p 0'; A SOO - 63.66 kPa (C) -;"(O:O'SV 0-81 Ay - 3 kip Use section having a length x. M 3.t - x2 El-:"TJ2v 3x - 0': 1';"- 153kPa Ans. - Use section having a length .x. Intensityof w - (-to\r L IM O: r at .x. -IS.\' 100 (t.x) [t(ii-.x1.x ] 153 kPa - 63.66 407.4 471 kPa 1:.:-c 0 1':. - 0 x2 D-82 A.y IS kip MA 100kip . ft M 0:.- 0 -0 -1 (- x4 0.5.r;3 - 2.25x) - 0'.- 0 M v EI' Moment about y axis: 0'; 0 since B is on neutral axis. Thus (f) L IM 0; -3x Integrate tWice. use v 0 at x O. v - 0 at x - 3 m - Torque: Tc -1(0:05)4 30(0.05)Tzz T Am. u.: Shearload: 'T;, 0 sinceat B. Q O. Momentabout.\' axis: Mc T(O.05)4 40(0.05) c 407.4kPa (C) 0'; T - O'c 0 (since C is on neutral axis) 0'\ c 0': c 0 Am. Ans. .\m. Ans. Ans. Am. - IS.x - 0.05.r3 - Integrate tWice, use v 0 at .\' 0, dvldx - -0 - O.OS .r3- 100 tfl11 IS.x EI "d.'t! v At' 100 0 at .x - 0 11 (2.5.0c3 - 0.0025r - SOr) .-\ns.
AP. D REVIEW FOR THE FUNDAMENTALS D-88 A - OF ENGINEERING 'Ir(0.02S)2- (0.015)2) 1.257(10-3) m2 1- 11'«0.02S)4 p :!!!:.!:!- c (KL)Z - 84.3kN -34 From Appendix C, consider distributed load and couple moment separately. woL3 ML 8A 45E/ 6Ei 4 (3)-' 20 (3) 45E/ - 12.4kip. ((2 1 0' D-89 p A ns. J -85 True Ans. - - (KL)Z - r (29 (loJ» (-1(0.5)4) [0.5(50)]2 225. k' 'P 0-90 P . -- ksi 36 ksi OK , 1 - 0,528 0.472 in. 3 Ans. - .,,229(10-') G ,4) (KL)2' 0.382in. P ". (0.382)2 3 A d 2r AM. - rz4)1 '[I (6)(12)f - . IT 0' Ans. 29( 10-')[7 (14 rz 0.528in, Ap 11' [(1)2 -40(0.52S)Z] 17.6 ,z m4 r (200(109»(262.04)(10-9) (0.5(5)]2 - .1!!.!:!-. (KL)Z' 40 -- r Ans. 2.03 kip (0.015)4) 267.04 (10-') 84.3(10-3) (1(}1) - 67.1 MPa 2SOMPa OK AP - 1.257 Thus -86 P 825 EXAMINATION [1 (40)]2 - 6.53ksi 36 ksi OK - 0.765 Ans.
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