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## Mechanical Design I

by: Princess Rolfson

56

0

19

# Mechanical Design I ME 371

Princess Rolfson
MSU
GPA 3.57

Staff

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COURSE
PROF.
Staff
TYPE
Class Notes
PAGES
19
WORDS
KARMA
25 ?

## Popular in Mechanical Engineering

This 19 page Class Notes was uploaded by Princess Rolfson on Saturday September 19, 2015. The Class Notes belongs to ME 371 at Michigan State University taught by Staff in Fall. Since its upload, it has received 56 views. For similar materials see /class/207532/me-371-michigan-state-university in Mechanical Engineering at Michigan State University.

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Date Created: 09/19/15
62 Use the method of joints to calculate the force in each member of the loaded truss illustrated and state Whether each member is in tension or compression Also calculate the reaction forces at A and C FIGURE P62 Solution The frame forms an equilateral triangle so that the angles y I I A Ax gto x 30 60quot AB Ac FIGURE S62 are all 60 A free body diagram of joint B yields BC cos 60 1000N AB cos 60 0 from the sum of the forces in the y direction and from the sum of the forces in the a direction we get AB cos 30 BC cos 30 0 AB BC Thus from the rst equation 1000 2AB cos 60 469 or AB 1000 139 tension and BC 1000 NCompression Moving to joint C ZED 0 yields CfE 1000 eos 30 or CE SGGN right The E F 0 implies 1000 eos 60 Mtension The equilibrium ofjoint A yields E E 0 implies Ax1000 cos 30 0r AB 866N left and SF 0 yields A 1000 eos 60 500l39 0 or A 1000l39 up 63 Use the method of joints to calculate the force in each member of the truss illustrated well the reaction forces at the pin B and roller FIGURE P63 Solution A free body diagram of each joint assuming tension in each member is illustrated y I I I y I A gt x I 500 y AB 45a Ac I I Ac I B BC 45 Bx pgtx AB BC ax FIGURE 363 The equilibrium equation for joint A in the a direction is 500 AC cos 45 0 or AC 707N compression The equilibrium equation for joint A in the y direction ins 707 cos 45 AB 0 471 or AB 500 N tension Next consider the joint at C implies BC 707eos 45 0 or BC 500 l39 tension E F 0 implies Cy 707 cos 45 0 or Cy 500K up Finally at point A BfB 500 0 or BfB 500139 left and By 500 0 or By 500N down 66 lalculate the force in each member of the truss illustrated well as the reaction forces of the pin at C and the roller at A b Add an additional 100 l39 force to the pin at D in the Vertical direction pointing downward and recalculate the forces in the members What changes 1000N B C 9 FIGURE P66 Solution The free body diagrams of each joint and of the entire body are rst constructed assuming each member to be in tension These are given in the gure 1000 1000 0y ac c Bclt olt cx 0 Bl x coseaI5 T oase35 1 4 I 5 AB I 4 B I I I I D 500 x case415 CD CD AB 6 A 005935 500 lt 9 ltgt gt 0 gt8 O M FIGURE 366 Note that the geometry is such that each angle is de ned by a 34 5 triangle The joint free body diagrams all contain more than two unknowns so starting with the FBD s of the whole body yields 2 MC 0 6i 43 X 500 M 3i gtlt 1000j 0 478 OI 0Ay12 200012 300012 0 or A 833 1 up 2 EE 0 C E 500 0 or CE 500 left 2F 0 Cy 1000833 0 or Cy 107 1 up Next consider the equilibrium equation for joint C ZED 0 BC CfE 0 or C E 500 N compression 2F 0 CD Cy 0 or Cy 16 N tension The FBD for D yields ZED 0 AB BD35 0 or AD 35BD 2F 0 CDBD45 0 or 54CD 209 N compression so that AD 125 N tension The FBD for A yields 21 0 500ADAB35 0 or AB 53625 1042 N compression 2F can be used a check and again yields Ay 833 N The FBD for B can be used as a check Alternately the matrix approach can be used In this case only the free body equations for joints A B C and D are needed From A 35AB AD 500 Z 0 lt45gtAB A 0 0 From B 35AB 35BD BC 0 45AB 45BD 1000 From C cm BC 0 3 cy CD 0 y 479 From D AD BD35 0 CD BD45 0 or in matrix form 35 1 0 0 0 0 0 0 AB 500 45 0 0 0 0 1 0 0 AD 0 35 0 1 35 0 0 0 0 BC 0 45 0 0 45 0 0 0 0 BD 7 1000 0 0 1 0 0 0 1 0 CD 0 0 0 0 0 1 0 0 1 Ay 0 0 1 0 35 0 0 0 0 CB 0 0 0 0 45 1 0 0 0 Cy 0 Solving Via Mathcad matrix inversion yields AB 1042 N compression Ay 833 N up AD 125 N tension CfE 500 N left b This is exactly the same part a except that the y equation for joint D becomes CD 45BD 100 0 Thus the matrix equation used in part A can be copied and the last element in the load vector changed The only Cy BC 500 N compression BD 208 N compression CD 1667 N tension Cy 1667 N up from 0 to 100 The new solution is AB 1042 N compression AD 125 N tension BC 500 N compression BD 208 N compression CD 267 N tension Ay 833 CfE 500 Nlc t Cy 267 elements that changed are the tension in CD and the reaction 480 67 A small walk way bridge carries the load indicated Calculate the forces in each member and whether they are in compression or in tension Also compute the reaction forces at the pin at A and the roller at point D 500 lb 1500 lb FIGURE P67 Solution First construct a free body diagram of each joint assuming each member is in tension Also make a FBD of the whole structure Starting with the FBD as a whole the equilibrium equation can be used to obtain the reaction forces directly ZM4oz5i x 500310 x 1500315 gtltDyj0 2500 15000 or Q T 1167 lb up ZFx03Am0 2F 0AyDy 200000rAy833lb up From joint A 2F 0 A AE sin 213 0 or AE 2 243 lb tension 2 EB 0 AB AE cos 218 0 or AB 2083 lb compression From joint E 2 EB 0 AE cos 218 EF 0 or EF 2 083 lb tension 2 F 0 BE AEsi11218 0 or BE 833 lb compression From joint D note 13 F and D have more than two unknown forces yet 1167 E F 21 137 2 J 0 D D s1n 8 0orD 51112180 3142 lb tension 481 1500 lb 500 lb y l 39 AB B Bc lax gt6 1 x o AB4 o X Bclt o gtcn 0tan39 25 218 e21s c BF AE Ay BE CF BE cF AE BF 1 BF 21 E X b4523 E EF F 500 1500 A Ax 5 5 5 gt ltU gt FIGURE 367 2 EB 0 DF cos 218 CD 0 CD 2918 lb compression From inspection of joint 0 CF 1500 lb compression and 130 2918 lb compression From joint F ZED 0 BFcos 218 EF DFcos 218 0 2083 3142 218 899 lb tension cos 218 Note the remaining equation at joint FZ F 0 and the two equations of equilibrium for joint 13 are not used as they are dependent but may now be used to check the previous calculations 482 lt x 632 A crane supports a 1000 N load Compute the forces in the members CE CF and DF A C E 05m G X 39B 05m E 05m 1000 N FIGURE P032 Solution Since a cut through the desired numbers can be made Without rst calculating the reactions at A and B only the freebody diagram of the section is required given in 8632 c c54 0395 G CF quot moon 45 DF FIGURE 3632 The three scalar equations of equilibrium are ZED0 CE DF CF 0 2F 0CF 10000 or CF14141 EMF 0 510005CE 0 or CE 1000 N and then from the equilibrium equation in the CCderCthl lI DF 2000 N 636 The owner of the storage shed wishes to hang a 250 kg engine from the roof illustrated in gure P035 and wants to ensure that the roof will not collapse due to the added weight of the motor The members are rated to hold tension and compression loads of up to 4000 N Use the method of sections to determine the loads in CE CF and DE and make a recommendation to the owner Solution First a freebody diagram of the crate yields the loads at points D and E This is illustrated in gure 3636a The equilibrium equations become 1000 N 1000 N 1 5 s 5 1 l TD TE 3 vquot 42 4 2m 1 45 l 1734 1734 739 250981 l Ax gt 7 7 7 7 7 a f 1 Av By M 1OUON ll 5 0E 2m DE 9 E 7m 45 1734 5 y n X FT 22265 C FIGURE 3636 1 1 F20z Ti T izo 2 Duff W2 F 0 T i1T71 2453 V0 E y W2 W2 2439 2 or TD TF 03f 1734 N A freebody diagram of the whole structure is rst required to obtain the reaction at B and this is given in Figure 8636b The moment equation about A yields EMA 51000 101000153y125i5jgtltma i j 10i 5i x1734 i j 0 or By 22265 1 as expected from symmetry ie 2000 24532 Next consider a free body diagram of the section to the right of a cut through the lines CE CF and CE illustrated in Figure 3636c The equations of equilibrium are note sinQZVIiQT cosQZVILE 5 1 FE02 CF CEi DE 1quot34 70 Z m lt gt95 2 1 F 202 1000 CE 7 1quot34 7 22263920 Z J 429 H a ZNIE 0 2CF522265 0 or CF 5566 N compression So that DE 4340 N tonsion Since both loads in DE and CF exceed the maximum member load allowed the garage owner should be advised against hanging the motor from the rafters 647 Compute the forces in members AC BC and BD The ower pot has a of 25 kg ETA 02m 0 02m E01m G 39 K 7 v FIGURE P647 Solution A free body of the entire structure is given in 64721 2E1 As Ba 0 EliA 0 Z Bat Z 0 Br 4906 2F A 2453 0 so that A 4906 A 2453 iv AK 6 3 25 981 Bx gt FIGURE 364721 A cut through the desired members is given in gure 647b The equilibrium equations are 2MB AC3 49053 0 or AC 4905 1 565 2453 A 49054 gtAc ac As I 2 BD B I 4905 gt 2 1 FIGURE 3647b 2 2 ZED 0 4905 BC V7T3 ED 4905 4905 0 5 3 1 F 0 24393BC 7BD7 0 Z J 0 M13 xS Solving yields BC 009 and BD 54834 N Note one could also just use the FBD of the removed section and avoid calculating the reactions 653 Calculate the force in member LK LJ and MK of gure P652 Solution The freebody diagram and resulting equilibrium equations for calculating the support forces is given in the solution to 652 The freebody diagram of a cut through LK is given in gure 8653 The equilibrium equation becomes 500 FIGURE 3653 7 7 7 5 7 V 7 ZFy 0LIXLIXEAIIXOOO O 2 1 ZED 0AL 0 XML05002MK0 so that MKz QSO lb LK0 LIZ 250 lb C 668 A can of paint 10 lb and a painter 180 lb are positioned on a step latter Compute the forces at each connection Model the step ladder as the 3 members shown with friction at support A and a frictionless connection at B The distance EC is 15 ft and CF is 1 ft FIGURE P668 Solution The exploded free body diagram is given in 8668 9 unknows A A By E Ex G Gm Hy fLE equations are 3 X 3 9 Solution from AG ZEBZ07 4BQBGBO 1 ZFy20AyCyGy0 2 EMA 0 C3 sin 75 Cy3 cos 75 G445 sin 75 Gy45 cos 75 0 from CE ZEDOac EE0 4 sz0 10 Cy Ey0 5 ZMC 0 10 3cos 75 Ey 0 6 612 cx 2x15cos75 Ex 4 FIGURE 3668 from BC ZED 0 G Ea 0 7 ZFy0 GyEy 180By0 8 ZMG 015 sin 750EE 15 cos 75 Ey 18021 cos 75 45 cos 75 By 0 9 In matrix form these 9 equations in 9 unknowns take the form a ex ex e es mgmmgmmmp I e G H W 1 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 0 1 0 0 0 3 sin 75 3 cos 75 0 0 45 sin 75 45 e05 5 0 0 0 1 0 1 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 3 cos 75 0 0 0 0 0 0 0 1 0 1 0 0 0 1 0 0 1 0 1 0 0 45 COS 75 0 0 15 sin 75 15 cos 75 0 0 0 0 0 0 10 10 0 180 18021COS75 which yields AE 0 CE 41264 lb Ey 12879 lb A 5896 lb Cy 22879 lb GB 41264 lb By 13104 lb Er 41264 lb G 36081 lb Note a quick check that A By 180 10 and AE 0 so that the a and y equilibrium equations of the rigid body are satis ed 614 675 lalculate the reaction forces at A and C and the forces at each connection FIGURE P675 The free body diagram of each member is given in the gure FIGURE 675 Note BC is a two force member There are 6 equations and 6 unknowns from the freebody diagram for AB 2E 0 yields 14 BfE 0 1 SF 0 yields A By 1000 0 2 EMA 0 yields 5By 4000 0 From BC ZED 0 yields Bm CD 0 4 2F 0 yields By Cy 0 5 2ch 0 yields 40 3C2E 0 6 628

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