hw7 - Problem 10.16 The cross-sectional area of the pris-...

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Unformatted text preview: Problem 10.16 The cross-sectional area of the pris- Free Body Diagram: matic bar is A : 2 in2 and the axial force P : 20 kip. ' Determine the normal and shear stresses on the plane P. Draw a diagram isolating the part of the bar to the right of plane P and show the stresses. 20 ,OOOI b {—— Solution: Thc area of plane P is: AP : (2 in2)/(cos 70°) AP : 5.8481n2 Summing vertical forces on the FBD: EFL, = D = —¢7Ap(sin 70°) + TAp(cos 70°) [1] ‘T : 2.75: Summing horizontal forces on the FBD: [2] SF3 z 0 : 20, 0001b ~— 0Ap(cos 70°) — TAP(SiI1700) Substituting Equation [1] into Equation [2]: 20, 000 lb = aAp{cos 70°)+(2.75U)AP (sin 70°) 2 cr(5.848 in2){cos 70° )+(2.750) (5.848 11:2) (sin 70°) ANS: or = 1169 psi Using the determined vaiue of 0' in Equation [1]: if) f} a “P .. 4 { ANS: "r = 2.750 = ~3215 psi Note: The negative sign con— forms to the sign convention given in Figure 3.15 T 2 2.75:: z —3215 psi r ProhleniV-ilfl'm Free Body Diagram: ‘1henormahstress’oh the plane P in Problem 1016 is 6000 psi, what is the axial force P? 6000A Eb re 20:. 70° J_ TA Solution: The area of plane P is: AP 2 (2 in2)/(cos 70°) AP = 5.848 in2 Summing vertical forces on the FBD: SF]; 2 0 = —0Ap(sin 70°) + TAP(CDS 70°) [1] 'r = 2.750' Summing horizontal forces on the FED: [2] SF“; : {J = P -u 044120305700)“ TAP(sin 70°) Suhtsituting Equation '[1] into Equation [2]: P : 0AP(cos 70°)+{2.750)(AP)(Sin 70°} : (60001b)(5.848in2)(003 70°)+(2.75)(60001b)(5.848 in2)(sin 70°) ANS: P = 102, 600 lb : 102.6 kip Problem 10.40 The bar has a circular cross section and moduius of elasticity E = 70 GPa. Parts A and G are 40 mm in diameter and part B is 80 mm in diameter. If F1 = 60 kN and F2 : 30 kN, what is the normal stress in part B? Free Body Diagram: «Mmmm—4hmmm4 é ZOOnnn ZOOWWH Solution: We must first determine the reactions at the left and right walls, We allow the right—hand side of the bar to “float.“The displacement of the “free” right-hand side of the bar is: _ FlLa __ FQLE i F3LA R _ AAEA ABEB AAEA 5 6R _ (60.000 N}(0.2 m) (30,000 N)(0.4 m} m 30,000 N)(0.2 m) . («WEN-roman? N/m“) (WW) (Toxiofl N/mfl) «W13 j(70x106 mm?) 63 = 0.0341 m = 34.1 mm The {caction at the right waii must be sufficient to prevent ANY dis- placement. lts magnitude is: 7 RE I . 0.4 0.2 m 0‘0341 m “ 70x105 lb/in (R (0.04 r1112) "(0.0531135 ) + (1' (0.01m? 4 RR = 6000 N The reaction at the left wali is: 253; :0:60,000N_6,000N~w30,000N7RL RL = 24,000N F The stress in section A is: —24,000 N+60.000 N I 50.08 m 08: = Note: The negative sign indicates a compressive stress. ANS: 0,4 = "7.16 MPa Problem 10.41 In Problem 10.40, if F1 = 60 kN, what force F2 will cause the normal stress in part 0 to be zero? Free Body Diagram: F1 |——+~fi400mmm-l 200mm Solution: For the normal stress in section C to be zero, we see that the displace- ment of the intersection of sections B and C must be zero. The equation for the displacement of the intersection of sections B and C‘ is: 530 =0: M_wm_— 71' ((0.04 uni/4) (70 x 109 N/mz) 71 ((0.04 1102/4} (70 x 109 N/mz) 11' ((0.08 1102/4} (70 x 109 mm?) ANS: F2 = 40, 000 N 2 40 kN Problem 14.9 Model the ladder rung as a simply sup- ported (pin-supported) beam and assume that the 200- 1b load exerted by the person‘s shoe is uniformly dis- tributed. Determine the internal forces an moment at A. Solution: The distributed load due to the weight exerted by the foot is: we = (200 1b)/{4 in) = 50 lb/in ' Draw the FED for the iadder rung and determine the reactions at the Eeft and right ends. 2001b |<——— 10in 5in——| E 2M1; = U = —[(50lb/in){4in)i(101n)+Rzy{15in) —} 3/9 = 133.311) T 25;; = o : —[(501b/in}(4in)]+Ry+Ly = —2001b+133.3lb+Ly —> Ly = 66.6711) T Since no forces are exerted in the as~directiom L3 : Rm : 0 Cut the beam through point A and draw the FBD: EUUIb lin |-* MA loin—4ifi5 | Ly EMA z 0 : m(65.67’1b}(10in)+£{501b/in)(2in)}{1in)+MA I ANS: MA : 567m in: SE}, = o = Ly—(EOIb/in)(2in)+VA = 66.67ib—1001b+VA ANS: VA : 33.31bT 251m : 0: PA ANS: PA =0 Problem 14.28 (a) Determine the internal forces and moment as functions of m. (b) Draw the shear force and bending moment diagrams. Solution: Draw the FBD and determine the reactions at point A. Summing vertical forces on the FED: EFy = o = — (100 Ib/ft) (6 m] + Ag Ag = 3001b Summing moments about point A: EMA : 0 = — (100 lb/ft) (6 ft)] (10 n) + MA MA =3000ft—1b ForU <9: <6ft: 10:0 2: V=—/Udm=0+01 atm=0,V=3001b 0 0'1 x 3001b V = 300 lb M=derszf300dw:300$+Cg o o a: : 0, M = m3, 000 ftelb 02 = —3, 000 n—sn ANS: M : (3000: — 3,000) ft—lb {1] Forfi < m < 12ft: w 2 (new) (we a) x 100) W + 100) d3; = $352+ 100w + 03 cc=12ft,V=0 03 = 0 ANS: V 2 ~33in + 100:1; 1: MAQ r—IUH / zj ,/’ I / 6Ft——i—— IGUlb/F‘t :n 2 M = dea; = f (—56ng + 100m) dm = (—359? + 50m2 +04) ft—lb 0 0 fromequationIl],a: = 6, M m —1.200ft-—1b0‘4 : m2,400 ftmlb ANS: M : (~%m3 + 503;2 ~ 2, 400) itwlb i E E g i E i E L Problem 14.55 The loads F = 200 N and C = 800 Nem. (a) Determine the internal forces and mo— ment as functions of w. (b) Draw the shear force and bending moment diagrams. Solution: Draw the FED for the beam and determine the reactions at points A and B. EMA = o = —(200N)(16m)+800N—m+By(8m) ——> 3,, = 3mm T EFy = D = ~200N+By+Ay m —200N+300N+Ay —> All = IDDN J, For0<rs<4mzw=0 V=——/Odm=0+01 at$=D,V=—IUUN,30:01=—-100N ANS: V = ~100 N |~4m+4m.|~—v~8m Mmedm=f(—1ODN)dm= (—100:z:+02) N—m atm=D,M=0, SO: 02 z 0 ANS: M= ~100a: N—m 11] For4m<nt<8rru 111:0 V:—ffldm=0+0’3atm=4,V=—EDDN, C3 = —100 N ANS: V: 7100N M: dex= f(—100N)d:c = (~100wN+C4) Nem from equation ii], at a: z 8, M = —160{) N—m, so: 04 = —800 ANS: M = (—1003 — 800) N—m For8m<m < 16m: 10:0 V:idecc:D+C'5ata::16,V:200N, so: 05 : 200 N ANS: V = 200 N M :demeOD N) do: : (200:1:+Cs) N—m ata: = 16, M : 0, so: 05 = —3,200 N—m ANS: M : (20033 e 3, 200) Nem i I t i Problem 15.9 The beam consists of material that will safely support a tensile or compressive stress of 350 MPa. Based on this criterion, determine the largest force F the beam will safely support if it has the cross section (a); if it has the cross section (b). (The two cross sections have approximately the same area.) Solution: The moment of inertia of the cross-section in case (a) is: W M3 7 (0.0233 m)(0.060 m)3 “T 12 m 12 The moment of inertia for the cross-section in case (b) is: : 4.194x10—7 m4 It: (0.050 m)(0.060 m)3 2 [(0.020 rn) (0.040 m)4 "' 12 12 Summing moments about point B to find the reaction at point A: 2MB = O : F(O.6 m) —— Ay(1.6 m) -—t A1, = 0.375F Maximum bending moment occurs at the point where the concentrated load is applied, so we calcuiate maximum bending stresses 1.0 m to the right of point A. Maximum bending moment is: MMAX 2 (0.375F)(1.0 m} 2 0.375F N—m For cross-section (a): 0.375F)(0.03 m) 195 N 2 ___ (— 350x /m 4.194x104 m4 ANS: F = 13.05 kN For cross-section (b): gamma N/mg 3 (0.375F)(0.03 m) 6.87x10—7 m4 ANS: F : 21.4 kN Problem 15.10 If the beam in Problem 15.9 is sub- jected to a force F = 6 kN, what is the maximum ten- sile stress due to bending at the cross section midway between the beam‘s supports in cases (a) and (b)? Solution: The moment of inertia of the cross~section in case (a) is: 3 I 2 I 3 fl : 91% z (00 33mg? 060m) :4'194X104 m4 The moment of inertia for the cross—section in case (b) is: It 2 (0.050 m)(0.000 m)3 _2 [(0.020 m) (0.040 1104] 12 12 Summing moments about point B to find the reaction at point A: 2MB = 0 = (0,000N)(0.6 m) — Ay{1.6m) —> 4., = 2250N T The bending moment at the midpoint of the beam is: M : (2256 N) (0.8 m) : 1800 Nam In case (a), the maximum bending stress is: (1800 N—m)(0.03 m) 4.194x10—7 m4 ANS: (a..)MAx : 128.8 MPa In case (b), the maximum bending stress is: (1800 N—m)(0.03 m) 6.87x 10"“r m4 ANS: (05mm = 78.6 MPa (Ja)MAX : (00)MAX = Free Body Diagram: _ 6.87x10"7m4 Free Body Diagram: l l l Problem 15.35 The beam whose cross section is shown consists of three planks of wood glued together; At a given axial position it is subjected to a shear force V = 2400 lb. What is the average shear stress at the neutral axis y’ : 0? Solution: Finding the centroid of the entire cross section (measuring from the TOP): |‘-4in.+-Ilintl-I-4in. I 7 _ m (2 in) (8 in) (4 in) + (2) [(2 in)(4 in)(7 111)} w 5 5 in (2 in)(8 in) + (2) [(2 in)(4 in)] ’ 8 :n. Calculating E: Calculating A': A' : {2 in)(5.5Inchc3) 2 11 in2 Calculating Q: Q = (—2.75 in)(11in2) = —30.25 in3 Calculating the moment of inertia: I = [W + (2 in)(8 in)_(4 in — 5.5 inf] +2 [ 12 12 Now calculating the average shear stress: _ V_Q _ (2,4001b)(—3a.25 i113) “VG _ bI ' (21n)(162.7in4) ANS: mm : 7223.11b/in2 Problem 15.36 In Problem 15.35, what are the mag- nitudes of the average shear stress acting on each glued joint? Solution: Finding the centroid of the entire cross section {measuring from the TOP): ' (2 in)(8 in) (4 in) + (2) [(2 in) (4 in) [7 in)} (2 in)(8 in) it (2) [(2 in)(4 in)] Calcuiating the moment of inertia: I _ [(2in)(8in)3 _ ' 12 r _ a: =5.5in + (2 in)(8 in)(4 in — 5.5 inf] +2 [ 12 Calculating E: if: 7in— 5.5m: 1.5m Calculating A’: A’ = (2 in)(4 in) = s in2 Calculating Q: Q 2 EA’ : (1.5 in)(B inz] : 12 1113 Now calculating the average shear stress: - VQ (2, 400 lb}(12 1:33) T = _ : __._ AVG 131 (2 in)(162.7 in4) ANS: TAVG m 88.51b/in2 (4 in)(2 in)3 (4 in)(2 in)3 J.- R‘fi 2m + (4 in}(2 in}(7 in — 5.511132] : 162.7in4 + (4 in)(2 in)(7 in v 5.5 i102] : 162.7in4 I Problem 15.46 The beam is subjected to a distrubuted load. For the cross section at m z 0.6 m, determine the average shear stress (a) at the neutral axis; (b) at y’ z 0.02 In. Free Body Diagram: 0.04 m 130 kN/rn M .. erlLfi—z Cross section. ’,a13U,UOUN/m 1 | Solution: Summing the moments about point B to determine the reaction at point A. 2 MB = 0 = — [gaseous N/m)(1.4 111)] (D.133m)+Ay(0.Bm) —> A5, =15,130N i Cut the beam at a: = 0.6 m and draw the FBD, (130,000N/1.4m)(x> / Summing the vertical fences to determine the shear force at :12 = 0.6 m: 1 :Fy = 0 z —15.130N [—§(55.714N/m)(0.6 m) +V a V = 31,844N A: (a) Using Equation (15-18) to determine the average shear stress at the neutral axis: 3V 3(—31,844 N T : EAT " 2(o.04 m){0.06 m) ANS: mm m —19.9 MPa (13) Using Equation (15-17) to determine the average stress at y’ : 0.02 m: _ 51/ a 2 , 2 _ 6(——31,844N) 0.06m 2 TAVG * m _ (3") i " (0.94 m)(0.06 m)3 2 ) m (0'02 my] ANS: mm = "11.06 MPa ...
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hw7 - Problem 10.16 The cross-sectional area of the pris-...

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