MEC430 ASSIGNMENT 1

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Toronto Metropolitan University *

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430

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Mechanical Engineering

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Oct 30, 2023

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Uploaded by PresidentMoon265

Department of Mechanical and Industrial Engineering Please select your current program below: ● Mechanical Engineering Industrial Engineering Course Number 430 Course Title MEC Semester/Year 5 th SEM 3 rd YEAR Section Number 6 Group Number 1 Assignment No. 1 Assignment Title Assigned questions Submission Date September 26th 2023 Due Date September 26th 2023 Student Name Student ID (xxxx1234) Signature * James Mohrhardt Xxxx97151 (Note: Remove the first 4 digits from your student ID) *By signing above you attest that you have contributed to this submission and confirm that all work you have contributed to this submission is your own work. Any suspicion of copying or plagiarism in this work will result in an investigation of Academic Misconduct and may result in a “0” on the work, an “F” in the course, or possibly more severe penalties, as well as a Disciplinary Notice on your academic record under the Student Code of Academic Conduct, which can be found online at: http://www.ryerson.ca/senate/policies/pol60.pdf .

↑ SFx 0 - HA 0 [Fy 0 > RA-()(2)(150) 100 Rx 0 R A R 350N/ ~ & [MA 0 -> (t)(2)(150) = (3)(150) (200) (200) Rx (250) 0 15000 40000 250Rp 0 ZOON zNmm RD 220NX HA q & d ↓ ↓ v ↓ D solving for RA with solutions : - A I 150mm Be-um 50mm RA 350 220 RA · I RD RA 130N al : 2Nmm (0 = X = 150mm) na a hi is - isoinnt I L X -> 130N e V , (x) 130 (t(x x) m , xx) (130)(x) (z)(rx)(x)() = 130 (i)()k) 130x () (s) ( =130 - 130 - =130x / 130(150) - : VA = 130 = -20N MA= ON m - =12000 N mm MB = 12000 N mm (t)(2)(150) N di trans - or A -A L - 130N V(x) 130 - (((2)(150) M2(x) (130)(x) ()(2)(1501( x -100) -- ION 130 x 150(x 100) .. VB V = 20N =15000 20x : MB 15000 20(150) =12000 N mm M , 15000 20(200) 11000 N mm

↑ R= <MA 0 (10kN) (1m) + (20kN)(2m) (20kw) (3m) (R=) (5m) = 10 40 40 5RE S Ho 5 RE 22 RA = 50-22 RA 28 kN V 28 kN Vi Vi P , Vi Vc 4 VE Vp = 22kN Shear diagram he Vi VA P2 18 20 = 2 20 = 28-18 =-2kN =-22kN 18V 18 ..... N : im < Im I -2 ams im i -22 ............... C S MA 0kw m [ MB Ma+(area ABC 46 ............ 0 (28 x 1) 44 = 28kN m/ 28 ..... i : " ! Mc Mi carea B to () A B C B ⑧ =28 (1x18) =46kN m/ Mp = Mc Carea 2 t0D) 46 (2x1) 44kN M/ M= Mp CareaDtoE) =42-(22 = 2) OKN MX

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[Fr 0 HA 0 [Fr 0 [FMA O -> + , VA V 8 + 18 MA 0 (8)(b) (1) (18)(9) V. (18) HA ↑ Un · VA + V 2b kips x = + (1)(18719) -> 2 - 67 - - - 18 ft aft GV = V , = 35 3 =14 33 Kips V , = 11 67 kips/ 14 33 8 33 --------- A ⑳ ⑳ SFA ki SF , = -8-(17 (18) 0 33 ..... -- C 3 D -267 . ------- ↑ SFB 43 -(1)(b) =- 11 667 kips 11 47 64 5 kips It =8 33 kiP SFx I (1) (9) -8 ---- son all parabolic Brin - =- 2 667 kips A BM ( )(4) (1)(b)(6) = 50kip It Bm, (2)(18) (81(12) (1) (15)(9) 0 Kip ft BM x (3)(9) (1) (9)(4 5) (8)(3) 64 5 kips It

Question 4 <MA 0 4R 4 - (2)(5) 1 R 1 5kN : RA 2 1 5 A Wm p dar RA 0 SKN HkN m ③ are Vx + 13x x 0 52x 1 5/10-4) R I ↑ Vx 0 5(x) + 1 5(x - 4: - xx - 47/ 0 5KN 1 . 5kN - Mx 1444 4 0 5(x7 1 . 5x 43 X 2 Mx 0 5[x + 1 . 54x 47 - 4 - 0 5xx 4 X 2 OkN --------- VA Ay 0 . 5 kN V V P VE Us' /area D - E) U SKN i " i VB VA 0 5kN 0 5+1 5 2 -(1)(2) · Vi V = Vp 0 5kw = 2kN 0kN 2 2m im E E ! I 1kW M -- ----- S MA 0 Mc Mi'+ /area Bec i ! 2 2m MB = MA + (Area A - B) - - -> -> = 3 + (1) (0 . 5) 22m 2 1m Im ⑳ = 0 (i)(k(2) --2 5kN m ⑧ 3kN m - - = IkN M Mi = MB + (couple) MD = Mc Carca <ep) = - 2 5 (0 5)(1) =1 - 4 = 2kW - M = 3kN -m Me = Mprarea of DeE) = z (I)(2)(2) =OkN m

Question 5 F (5))9)(3) , = I 13 5 Ax & rand at a 3 L -- X S E 2 Fx Ay 0 Si =MA 13 5(3) By(4) V(x) Ayxx a3 + (x - a P o (x ab + Bycx - a) 2b By = b 75 V(x) = 6 752 =7 % + 5(x2 3x + 75(x - 3 · S Fy Ay54 75 - 13 5 M ! Ay = b 75- 13 5 3 Ay b 75 Mix Ayx- al+ x-ay -Ex-a Byx- ac M(x) = 4 , 75xx3 ;y<x7 - 1 . 5(x) + 4 75xx - x shear : 6 75- V(0-1 0 1 5 = one v(0 + 6 75 5 25 ↑ ~(5) 6 15 + (5)(6) - 3(b) ↓ 9 2 67 5 . 25 V(6) 4 - 35 + (5)(4)2 3(6) + 4 75 1 5 V(a) b 75 + =(9) - B)(9 + 6 - 75 =0 Moment : 8 . 39 ⑧ V(x) 0 4 75(x (3))x) + I(x)2 0 (5)(x) - 3x + 4 75 x 263 ⑳ ⑧ M101 0 kip-fr ⑳ -1 5 - -- ---- M(7 44) 6 752 34) 1 - 5(2 + 41 + 58(3 6473 =5 39 kip It M(x) b , 15(4) + (6 - 15(b)2 =-1 5 kiPET M(a) = 6 . 75/9) + (a) - s . 519)" +415/3) = 0 Ki

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Question 6 F 1 5/24) 36kN = 34kN()(24)(1) Ax ↓ ↓ 1 12kn -> ↑ Ay ↑ Dy ⑲ [Fx Ax 0 S : (MA 0 - 36(1) 36((1 75) 12(2 87) + By /3 5 Vix= Ay ex-as' -P , -aso-PraxaY' x- 94'] + (x - a xx a ⑮ P = 36 36(11 . 751 - 1212 83) - 13 5 +Dy <x a) Dy 38kN vex 466x 028 34(x - 18 - 24((x x x 2 517 + 12 25[Fy 0 Ax 36 - 36 12 + 38 (x 3 53 + 38xx - 3 57 Ay 46kN/ M : Mix = Ayax-as-P . cras' [x a) + x a] + [ - a xx - a)) Dy x a) M(x) = 46ax 342x k - 1(x 13 + xx - 2 . 5323 + 4)(x - 2 523 - 4x 3 537 + 38(x 3 . 5) 46 - V(0) = 0 V(3 54 = 44 - 33 24(2 . 5 - 1 12 1 10 I V(0 + ) 46(07 - 36(098 = O - · I 10 kN ...... 16 - 1 42 I I S V(2 5) = 46 - 36 - 24)1 5) 110 24(1) b 8 I 1 b 10 I S 7 = 26kN d I I i 0 - 24(x) + 34 I I ~(3 . 5) 44 34 24(2 . 5-1) 1 x 1 42m i ! = 38kN ! M101 0 kN M ! : M(1) = 44kw m M(1 22) 44(1 431-36(0 421 120 . 4272 i 48 00kw M M(2 57 43(2 57 3491 5) - 12/1 52 = 34kN m M13 5) = 46(3 5) 36(2 5) R(2 54 + 4(1)2 -0

Question 7 = (120)(4) SMA 3014) 30(8) + Py(10) =GOlD 50 16 dy 64/b At ↓ ↓ 2 Fy An -60 -50+64 ↑ ↑ Dy At = 441b Ay shear section VixI Ay ax-os-G-o- -- a] + Pox-a' <x-a +DyCx-a V(x 44(x30 - I /xx 02 xx - xy + 20(x - x1 50(x 8) + 64xx 10 Moment section : Mix = Ayeran' - - [ax-a > <x-ax] <x - " - x a + By(x - a) M(x 44(x3 - ((x3 xx x3) + 10(x - 4)" - 50xx 8) + 44cx 1 46 - · rol-uncal" - co ! V(0) =0 46 = 441b S S 5 25 v (b) 44 - =(i) S x = 5 25 ↓ =- 14 Ib I -14 V 18 ) 44 =182 2) + 20(w) 64 =-141b V(84) 4 =(2 - 7 + 40 50 = -41b V(15 44 =(100 161 + 80 50 1956 = ⑧ ⑧ = 621b 128 ⑧ 0 8 . 25 M(0) 01b . in 128 - M15 25) (16(5 . 25) - =15 . 7513 154 - =161/bin +6 - M(6) 44(4) - 5(6) = 156 Ib in m(0) 46(8) -183 2) 10(2) : =120 Thin m(10) = 44(10) - 5(10 -4) + (r(14) - 100 = 0 Ibin

Question 8 Ar , disangan ↑ (MA 0 cy(61 1 5(1) 4(1 5) A = 1 75kN -> Ay <F Ay 3 1 5 1 75 Cy Shear section : (x = An(x as Pr(<x - 0xx - x) + f((x - a - (x - x") P , cr x (y(x - x Vix 5 75(x70 3/(x) - xx 3) + b((x (x x2) - (x - x" 1 75(x - 3 moment section : m(x 5 75xx-02 (x 02 (x x") + [x - 03 1 - x3) <x x + 1 15cx x V 10-1 0 KN intersection 3 75- (20%) 5 75 kN V(x) 0 5 . 25 3x 6x2 ~ 2 Im (13) 5 , 75-3(3) + ! (9) root 2 18 X S · ⑳ =-1 75kN v(x) 5 75 -3(3) + 5(34 9) - (3) + 1 75 S OKN 1 75- ⑧ S I S S I I I M101 0KN m I 5 , 98 - ⑧ I I M(2-18) = 5 . 25 (1 181 += 12 18- 12 185 5 25 - I ' ⑧ I = 5 . 98kN -m ! M(3) = 5 75(3) + =(3) - (3) ⑳ 5 - 35kW m 5 25 5 90 M(h) = 5 - 75(6) - ( - 34+y(63 3) - 12 =0 KN

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