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Problem 6.1 In Active Example 6.1, suppose that in addition to the 2-kN downward force acting at point D, a 2-kN downward force acts at point C. Draw a sketch of the truss showing the new loading. Determine the axial forces in members AB and AC of the truss. C 5 m5 m A D B 2 kN 3 m 3 m Solution: The new sketch, a free-body diagram of the entire truss and a free-body diagram of the joint at A are shown. The angle ˛ between CD and BD is ˛ D tan�1�6/10� D 31.0° Using the entire truss, the equilibrium equations are Fx : Ax C B D 0 Fy : Ay � 2 kN� 2 kN D 0 MA : ��2 kN��5 m�� �2 kN��10 m� C B�6 m� D 0 Solving yields Ax D �5 kN, Ay D 4 kN, B D 5 kN Using the free-body diagram of joint A, the equilibrium equations are: Fx : Ax C TAC cos ˛ D 0 Fy : Ay � TAB � TAC sin ˛ D 0 Solving yields TAB D 1 kN, TAC D 5.83 kN Because both values are positive, we know that both are in tension AB : 1 kN (T), AC : 5.83 kN (T) 386 c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.2 Determine the axial forces in the members of the truss and indicate whether they are in tension (T) or compression (C). C 800 N 0.7 m 0.7 m A B 0.4 m 20� Solution: We start at joint A ∑ Fx : � 7p 65 FAB C 7p 65 FAC � �800 N� sin 20° D 0 ∑ Fy : � 4p 65 FAB � 4p 65 FAC � �800 N� cos 20° D 0 Solving we have FAB D �915 N, FAC D �600 N 7 7 44 800 N A FACFAB 20° Next we move to joint C ∑ Fx : � 7p 65 FAC � FBC D 0) FBC D 521 N C Cy FAC FCB 7 4 In summary we have FAB D 915 N�C�, FAC D 600 N�C�, FBC D 521 N�T� Problem 6.3 Member AB of the truss is subjected to a 1000-lb tensile force. Determine the weight W and the axial force in member AC. A B W C 60 in 60 in 60 in Solution: Using joint A ∑ Fx : � 2p 5 �1000 lb�� 1p 2 FAC D 0 ∑ Fy : � 1p 5 �1000 lb�� 1p 2 FAC �W D 0 Solving we have FAC D �1265 lb, W D 447 lb In summary we have W D 447 lb, FAC D 1265 lb�C� 1000 lb A2 1 1 1 FAC W c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 387 Problem 6.4 Determine the axial forces in members BC and CD of the truss. 600 lb D E 3 ft 3 ft 3 ft 3 ft A C B Solution: The free-body diagrams for joints E, D, and C are shown. The angle ˛ is ˛ D tan�1�3/4� D 36.9° Using Joint E, we have Fx : ��600 lb�� TCE sin ˛ D 0 Fy : �TCE cos ˛� TDE D 0 Using Joint D, we have Fx : �TCD � TBD sin ˛ D 0 Fy : TDE � TBD cos ˛ D 0 Finally, using Joint C, we have Fx : TCD C TCD sin ˛� TAC sin ˛ D 0 Fy : TCE cos ˛� TAC cos ˛� TBC D 0 Solving these six equations yields TCE D �1000 lb, TDE D 800 lb TCD D �600 lb, TAC D �2000 lb TBC D 800 lb, TBD D 1000 lb A positive value means tension and a negative value means compres- sion Thus BC : 800 lb (T), CD : 600 lb (C) 388 c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.5 Each suspended weight has mass m D 20 kg. Determine the axial forces in the members of the truss and indicate whether they are in tension (T) or compression (C). 0.32 m0.16 m0.16 m 0.4 m A B C D m m Solution: Assume all bars are in tension. Start with joint D ∑ Fy : 5p 61 TAD � 196.2 N D 0 ∑ Fx : � 6p 61 TAD � TCD D 0 Solving: TAD D 306 N, TCD D �235 N TAD TCD 5 6 196.2 N D Now work with joint C ∑ Fy : 5p 29 TAC � 196.2 N D 0 ∑ Fx : � 2p 29 TAC � TBC C TCD D 0 Solving: TAC D 211 N, TBC D �313 N TAC 5 2 196.2 N C TBC TCD Finally work with joint A ∑ Fy : � 5p 29 �TAB C TAC�� 5p 61 TAD D 0 ) TAB D �423 N T TAB TAC TAD A 2 2 5 5 5 6 In summary: TAB D 423 N�C� TAC D 211 N�T� TAD D 306 N�T� TBC D 314 N�C� TCD D 235 N�C� c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 389 Problem 6.6 Determine the largest tensile and com- pressive forces that occur in the members of the truss, and indicate the members in which they occur if (a) the dimension h D 0.1 m; (b) the dimension h D 0.5 m. Observe how a simple change in design affects the maximum axial loads. 0.6 m 0.4 m h 1.2 m 0.7 m1 kN C D B A Solution: To get the force components we use equations of the form TPQ D TPQePQ D TPQXiC TPQYj where P and Q take on the designations A, B, C, and D as needed. Equilibrium yields At joint A: ∑ Fx D TABX C TACX D 0, and ∑ Fy D TABY C TACY � 1 kN D 0. At joint B: ∑ Fx D �TABX C TBCX C TBDX D 0, and ∑ Fy D �TABY C TBCY C TBDY D 0. At joint C: ∑ Fx D �TBCX � TACX C TCDX D 0, and ∑ Fy D �TBCY � TACY C TCDY C CY D 0. At joint D: ∑ Fx D �TCDX � TBDX C DX D 0, and ∑ Fy D �TCDY � TBDY C DY D 0. Solve simultaneously to get TAB D TBD D 2.43 kN, TAC D �2.78 kN, TBC D 0, TCD D �2.88 kN. Note that with appropriate changes in the designation of points, the forces here are the same as those in Problem 6.4. This can be explained by noting from the unit vectors that AB and BC are parallel. Also note that in this configuration, BC carries no load. This geometry is the same as in Problem 6.4 except for the joint at B and member BC which carries no load. Remember member BC in this geometry — we will encounter things like it again, will give it a special name, and will learn to recognize it on sight. 0.6 m 1.2 m CY DY DX TBC −TBC TBD TCD TAB TAC −TAB −TAC−TCD −TBD B y h C D A x 1 kN 0.4 m 0.7 m (b) For this part of the problem, we set h D 0.5 m. The unit vectors change because h is involved in the coordinates of point B. The new unit vectors are eAB D �0.986iC 0.164j, eAC D �0.864i� 0.504j, eBC D 0i� 1j, eBD D �0.768i� 0.640j, and eCD D �0.832iC 0.555j. We get the force components as above, and the equilibrium forces at the joints remain the same. Solving the equilibrium equations simul- taneously for this situation yields TAB D 1.35 kN, TAC D �1.54 kN, TBC D �1.33, TBD D 1.74 kN, and TCD D �1.60 kN. These numbers differ significantly from (a). Most significantly, member BD is now carrying a compressive load and this has reduced the loads in all members except member BD. “Sharing the load” among more members seems to have worked in this case. 390 c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.7 This steel truss bridge is in the Gallatin National Forest south of Bozeman, Montana. Suppose that one of the tandem trusses supporting the bridge is loaded as shown. Determine the axial forces in members AB, BC, BD, and BE. 17 ft 17 ft 17 ft 17 ft A B D F H GEC 8 ft 10 kip 10 kip 10 kip Solution: We start with the entire structure in order to find the reaction at A. We have to assume that either A or H is really a roller instead of a pinned support. ∑ MH : �10 kip��17 ft�C �10 kip��34 ft�C �10 kip��51 ft� � A�68 ft� D 0) A D 15 kip 17 ft A 10 kip 10 kip 10 kip H 17 ft 17 ft 17 ft Now we examine joint A ∑ Fy : 8p 353 FAB C A D 0) FAB D �35.2 kip 17 8 A FAB FAC Now work with joint C ∑ Fy : FBC � 10 kip D 0) FBC D 10 kip FAC FCE FBC C 10 kip Finally work with joint B ∑ Fx : � 17p 353 FAB C 17p353 FBE C FBD D 0 ∑ Fy : � 8p 353 FAB � 8p 353 FBE � FBC D 0 Solving we find FBD D �42.5 kip, FBE D 11.74 kip 17 8 17 8 FBE FBC FAB B FBD In Summary we have FAB D 35.2 kip�C�, FBC D 10 kip�T�, FBD D 42.5 kip�C�, FBE D 11.74 kip�T� c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 391 Problem 6.8 For the bridge truss in Problem 6.7, determine the largest tensile and compressive forces that occur in the members, and indicate the members in which they occur. Solution: Continuing the solution to Problem 6.7 will show the largest tensile and compressive forces that occur in the structure. Examining joint A we have ∑ Fx : 17p 353 FAB C FAC D 0) FAC D 31.9 kip Examining joint C ∑ Fx : �FAC C FCE D 0) FCE D 31.9 kip Examining joint D ∑ Fy : �FDE D 0) FDE D 0 DFBD FDF FDE The forces in the rest of the members are found by symmetry. We have FAB D FFH D 35.2 kip�C� FAC D FGH D 31.9 kip�T� FBC D FFG D 10 kip�T� FBD D FDF D 42.5 kip�C� FBE D FEF D 11.74 kip�T� FCE D FEG D 31.9 kip�T� FDE D 0 The largest tension and compression members are then FAC D FEG D FCE D FGH D 31.9 kip�T� FBD D FDH D 42.5 kip�C� 392 c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.9 The trusses supporting the bridge in Problems 6.7 and 6.8 are called Pratt trusses. Suppose that the bridge designers had decided to use the truss shown instead, which is called a Howe truss. Determine the largest tensile and compressive forces that occur in the members, and indicate the members in which they occur. Compare your answers to the answers to Problem 6.8. 17 ft 17 ft 17 ft 17 ft A B D F H GEC 8 ft 10 kip 10 kip 10 kip Solution: We start with the entire structure in order to find the reaction at A. We have to assume that either A or H is really a roller instead of a pinned support. ∑ MH : �10 kip��17 ft�C �10 kip��34 ft�C �10 kip��51 ft� � A�68 ft� D 0) A D 15 kip A H10 kips 10 kips 10 kips Now we examine joint A ∑ Fy : 8p 353 FAB C A D 0) FAB D �35.2 kip ∑ Fx : 17p 353 FAB C FAC D 0) FAC D 31.9 kip A FAC FAB 17 8 Now work with joint B ∑ Fx : � 17p 353 FAB C FBD D 0) FBD D �31.9 kip ∑ Fy : � 8p 353 FAB � FBC D 0) FBC D 15 kip FBD FBC FAB B 17 8 Next work with joint C ∑ Fy : FBC C 8p 353 FCD � 10 kip D 0) FCD D �11.74 kip ∑ Fx : FCE C 17p 353 FCD � FAC D 0) FCE D 42.5 kip FCD FBC FCE 10 kip FAC C 17 8 Finally from joint E we find ∑ Fy : FDE � 10 kip D 0) FDE D 10 kip E FCE FDE FEG 10 kip The forces in the rest of the members are found by symmetry. We have FAB D FFH D 35.2 kip�C� FAC D FGH D 31.9 kip�T� FBD D FDF D 31.9 kip�C� FBC D FFG D 15 kip�T� FCD D FDG D 11.74 kip�C� FCE D FEG D 42.5 kip�T� FDE D 10 kip�T� The largest tension and compression members are then FCE D FEG D 42.5 kip�T� FAB D FFH D 35.2 kip�C� c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 393 Problem 6.10 Determine the axial forces in members BD, CD, and CE of the truss. F 400 mm 400 mm 400 mm 300 mm 300 mm 6 kN A B C D E G Solution: The free-body diagrams of the entire truss and of joints A, B, and C are shown. The angle ˛ D tan�1�3/4� D 36.9° From the free-body diagram of the entire truss Fy : Ay � 6 kN D 0 MG : �6 kN��400 mm�C Ax�600 mm� � Ay�1200 mm� D 0 Solving, Ax D 8 kN, Ay D 6 kN Using joint A, Fx : Ax C TAB C TAC cos ˛ D 0 Fy : Ay C TAC sin ˛ D 0 Solving we find TAB D 0, TAC D �10 kN Because joint B consists of three members, two of which are parallel, and is subjected to no external load, we can recognize that TBD D TAB D 0 and TBD D 0 Finally we examine joint C Fx : TCE C TCD cos ˛� TAC cos ˛ D 0 Fy : �TAC sin ˛� TCD sin ˛� TBC D 0 } ) TCD D 10 kN, TCE D �16 kN In summary BD : 0, CD : 10 kN (T), CE : 16 kN (C) 394 c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.11 The loads F1 D F2 D 8 kN. Determine the axial forces in members BD, BE, and BG. 3 m A B D E G F2 F1 4 m C 4 m 3 m Solution: First find the external support loads and then use the method of joints to solve for the required unknown forces. (Assume all unknown forces in members are tensions). External loads: y x B A E G GY C D AX AY F1 = 8 kN F2 = 8 kN 3 m 8 m 3 m ∑ Fx : Ax C F1 C F2 D 0 (kN) ∑ Fy : Ay CGy D 0 C ∑ MA : 8Gy � 3F2 � 6F1 D 0 Solving for the external loads, we get Ax D �16 kN �to the left� Ay D �9 kN �downward� Gy D 9 kN �upward� Now use the method of joints to determine BD, BE, and BG. Start with joint D. Joint D : BD DE D x y F1 = 8 kN θ cos � D 0.8 sin � D 0.6 � D 36.87° ∑ Fx : F1 � BD cos � D 0 ∑ Fy : � BD sin � � DE D 0 Solving, BD D 10 kN �T� DE D �6 kN �C� Joint E : BE DE EG x y F2 = 8 kN DE D �6 kN ∑ Fx D DE� EG D 0 ∑ Fy D �BEC F2 D 0 Solving: EG D �6 kN �C� BE D 8 kN �T� c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 395 6.11 (Continued ) Joint G : EG y x CG BG GY θ �EG D �6 kN �C�� Gy D 9 kN ∑ Fx : �CG� BG cos � D 0 ∑ Fy : BG sin � C EGCGy D 0 Solving, we get BG D �5 kN �C� CG D 4 kN �T� Thus, we have BD D 10 kN �T� BE D 8 kN �T� BG D �5 kN �C� Problem 6.12 Determine the largest tensile and compressive forces that occur in the members of the truss, and indicate the members in which they occur if (a) the dimension h D 5 in; (b) the dimension h D 10 in. Observe how a simple change in design affects the maximum axial loads. 20 in 20 in 20 in 30� 800 lb A B CE D h Solution: Starting at joint A ∑ Fx : � 20p h2 C 202 FAB � FAC C �800 lb� sin 30 ° D 0 ∑ Fy : hp h2 C 202 FAB � �800 lb� cos 30 ° D 0 800 lb A 20 h FAB FAC Next joint B ∑ Fx : �FBD � 20p h2 C 202 FBC C 20p h2 C 202 FAB D 0 ∑ Fy : � hp h2 C 202 FBC � hp h2 C 202 FAB D 0 20 h h B FBD FABFBC 20 Finally joint C ∑ Fx : � 20p h2 C 202 FCD C 20p h2 C 202 FBC � FCE C FAC D 0 ∑ Fy : hp h2 C 202 FCD C hp h2 C 202 FBC D 0 2020 h h FCD FBC C FACFCE (a) Using h D 5 in we find: FAB D 2860 lb�T�, FAC D 2370 lb�C�, FBD D 5540 lb�T� FBC D 2860 lb�C�, FCD D 2860 lb�T�, FCE D 7910 lb�C� ) FBD D 5540 lb�T� FCE D 7910 lb�C� (b) Using h D 10 in we find: FAB D 1549 lb�T�, FAC D 986 lb�C�, FBD D 2770 lb�T� FBC D 1549 lb�C�, FCD D 1549 lb�T�, FCE D 3760 lb�C� ) FBD D 2770 lb�T� FCE D 3760 lb�C� 396 c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.13 The truss supports loads at C and E. If F D 3 kN, what are the axial forces in members BC and BE? A B C D E G 1 m F 2F 1 m 1 m 1 m Solution: The moment about A is ∑ MA D �1F� 4FC 3G D 0, from which G D 5 3 F D 5 kN. The sums of forces: ∑ FY D AY � 3FC G D 0, from which AY D 4 3 F D 4 kN. ∑ FX D AX D 0, fromwhich AX D 0. The interior angles GDE, EBC are 45°, from which sin ˛ D cos ˛ D 1p 2 . Denote the axial force in a member joining I, K by IK. (1) Joint G : ∑ Fy D DGp 2 CG D 0, from which DG D �p2G D � 5 p 2 3 F D �5p2 kN �C�. ∑ Fx D �DGp 2 � EG D 0, from which EG D �DGp 2 D 5 3 F D 5kN �T�. (2) Joint D : ∑ Fy D �DE� DGp 2 D 0, from which DE D 5 3 F D 5 kN �T�. ∑ Fx D �BDC DGp 2 D 0, 1 m 1 m 1 m 1 m AY AY AX F G2F DG DE BD DG EG AC AB AC BC CE F G Joint G Joint A Joint C Joint D Joint E 45° 45° 45° 45° 45° CE EG DE BE from which BD D � 5 3 F D �5 kN �C�. (3) Joint E : ∑ Fy D BEp 2 � 2FC DE D 0, from which BE D 2p2F�p2DE D p 2 3 F D p2 kN �T�. ∑ Fx D �CE� BEp 2 C EG D 0, from which CE D EG� BEp 2 D 4 3 F D 4 kN �T�. c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 397 6.13 (Continued ) (4) Joint A: ∑ Fy D Ay � ACp 2 D 0, from which AC D 4 p 2 3 F D 4p2 kN �T�. ∑ Fx D ABC ACp 2 D 0, from which AB D �4 3 F D �4 kN �C�. (5) Joint C : ∑ Fy D BCC ACp 2 � F D 0, from which BC D F� ACp 2 D � 1 3 F D �1 kN �C�. Problem 6.14 If you don’t want the members of the truss to be subjected to an axial load (tension or compres- sion) greater than 20 kN, what is the largest acceptable magnitude of the downward force F? 12 m 3 m A F C D B 4 m Solution: Start with joint A ∑ Fx : �FAB cos 36.9° � FAC sin 30.5° D 0 ∑ Fy : �FAB sin 36.9° � FAC cos 30.5° � F D 0 A 36.9° 30.5° F FAB FAC Now work with joint C ∑ Fx : �FCD � FBC sin 36.9° C FAC sin 30.5° D 0 ∑ Fy : FBC cos 36.9° C FAC cos 30.5° D 0 36.9° 30.5° CFCD FBC FAC Finally examine joint D ∑ Fy : FBD D 0 FBD DDx FCD Solving we find FAB D 1.32F, FAC D �2.08F, FCD D �2.4F, FBC D 2.24F, FBD D 0 The critical member is CD. Thus 2.4F D 20 kN) F D 8.33 kN 398 c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.15 The truss is a preliminary design for a structure to attach one end of a stretcher to a rescue helicopter. Based on dynamic simulations, the design engineer estimates that the downward forces the stretcher will exert will be no greater than 1.6 kN at A and at B. What are the resulting axial forces in members CF, DF, and FG? 300 mm 290 mm 390 mm 200 mm 480 mm 150 mm AB D C G F E Solution: Start with joint C ∑ Fy : 48p 3825 FCF � 1.6 kN D 0) FCF D 2.06 kN FCF 39 48 C 1.6 kN FCD Now use joint F ∑ Fx : � 59p 3706 FFG � 29p 3145 FDF C 39p 3825 FCF D 0 ∑ Fy : 15p 3706 FFG � 48p 3145 FDF � 48p 3825 FCF D 0 Solving we find FDF D �1.286 kN, FCF D 2.03 kN FDF FCF FFG 59 15 F 39 48 48 29 In Summary FCF D 2.06 kN�T�, FDF D 1.29 kN�C�, FCF D 2.03 kN�T� c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 399 Problem 6.16 Upon learning of an upgrade in the heli- copter’s engine, the engineer designing the truss does new simulations and concludes that the downward forces the stretcher will exert at A and at B may be as large as 1.8 kN. What are the resulting axial forces in members DE, DF, and DG? Solution: Assume all bars are in tension. Start at joint C ∑ Fy : 16p 425 TCF � 1.8 kN D 0) TCF D 2.32 kN ∑ Fx : � 13p 425 TCF � TCD D 0) TCD D �1.463 kN C TCF TCD 13 16 1.8 kN Next work with joint F ∑ Fx : � 59p 3706 TFG � 29p 3145 TDF C 13p 425 TCF D 0 ∑ Fy : 15p 3706 TFG � 48p 3145 TDF � 48p 425 TCF D 0 Solving TDF D �5.09 kN, TFG D 4.23 kN TFG F TDF TCF 29 48 13 16 15 59 Next work with joint B ∑ Fx : � 3p 13 TBE D 0) TBE D 0 ∑ Fy : 2p 13 TBE C TBD � 1.8 kN D 0) TBD D 1.8 kN B TBDTBE 3 2 1.8 kN Finally work with joint D ∑ Fx : �TDE � 10p 541 TDG C 29p 3145 TDF C TCD D 0 ∑ Fy : 21p 541 TDG C 48p 3145 TDF � TBD D 0 Solving: TDG D 6.82 kN, TDE D �7.03 kN TDE TCD TBD TDG D TDF 21 10 48 29 In summary: TDE D 7.03 kN�C�, TDF D 5.09 kN�C�, TDG D 6.82 kN�T� 400 c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.17 Determine the axial forces in the members in terms of the weight W. A B E D C 1 m 1 m 0.8 m 0.8 m 0.8 m W Solution: Denote the axial force in a member joining two points I, K by IK. The angle between member DE and the positive x axis is ˛ D tan�1 0.8 D 38.66°. The angle formed by member DB with the positive x axis is 90° C ˛. The angle formed by member AB with the positive x axis is ˛. Joint E : ∑ Fy D �DE cos ˛�W D 0, from which DE D �1.28W �C� . ∑ Fy D �BE� DE sin ˛ D 0, from which BE D 0.8W �T� Joint D : ∑ Fx D DE cos ˛C BD cos ˛�CD cos ˛ D 0, from which BD �CD D �DE. ∑ Fy D �BD sin ˛C DE sin ˛�CD sin ˛ D 0, from which BD CCD D DE. Solving these two equations in two unknowns: CD D DE D �1.28W �C� , BD D 0 Joint B : ∑ Fx D BE� AB sin ˛� BD sin ˛ D 0, from which AB D BE sin ˛ D 1.28W�T� ∑ Fy D �AB cos ˛� BC D 0, from which BC D �AB cos ˛ D �W�C� c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 401 Problem 6.18 The lengths of the members of the truss are shown. The mass of the suspended crate is 900 kg. Determine the axial forces in the members. 12 m 12 m 5 m 13 m 13 m C D B A 40� Solution: Start with joint A ∑ Fx :� FAB cos 40° � FAC sin 27.4° D 0 ∑ Fy :� FAB sin 40° � FAC cos 27.4° � �900 kg��9.81 m/s2� D 0 A FAC FAB 8829 N 40° 27.4° Next work with joint C ∑ Fx :� FCD cos 40° � FBC cos 50° C FAC sin 27.4° D 0 ∑ Fy :� FCD sin 40° C FBC sin 50° C FAC cos 27.4° D 0 27.4° 50° 40° FAC FCD C FBC Finally work with joint B ∑ Fy : FAB cos 50° � FBC sin 50° � FBD cos 27.4° D 0 50° 50° 27.4° FAB FBC FBD T B Solving we find FAB D 10.56 kN D 10.56 kN�T� FAC D �17.58 kN D 17.58 kN�C� FCD D �16.23 kN D 16.23 kN�C� FBC D 6.76 kN D 6.76 kN�T� FBD D 1.807 kN D 1.807 kN�T� 402 c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.19 The loads F1 D 600 lb and F2 D 300 lb. Determine the axial forces in members AE, BD, and CD. F2 F1 D A B C 4 ft 3 ft G E 6 ft 4 ft Solution: The reaction at E is determined by the sum of the moments about G: MG D C6E� 4F1 � 8F2 D 0, from which E D 4F1 C 8F2 6 D 800 lb. The interior angle EAG is ˛ D tan�1 ( 6 8 ) D 36.87°. From similar triangles this is also the value of the interior angles ACB, CBD, and CGD. Method of joints: Denote the axial force in a member joining two points I, K by IK. Joint E : ∑ Fy D EC AE D 0, from which AE D �E D �800 lb �C� . ∑ Fy D EG D 0, from which EG D 0. Joint A: ∑ Fy D �AE� AC cos ˛ D 0, from which AC D �AE 0.8 D 1000 lb�T�. ∑ Fy D AC sin ˛C AB D 0, from which AB D �AC�0.6� D �600 lb�C�. Joint B : ∑ Fy D BD sin ˛� AB� F1 D 0, GX GY 6 ft 4 ft 4 ft F1 F2 E EG E AE AE AC AB BD BC AB DG CD F2 F1 BD α αα Joint E JointA Joint B Joint D from which BD D F2 C AB 0.6 D �300 0.6 D �500 lb�C� . ∑ Fx D �BC� BD cos ˛ D 0, from which BC D �BD�0.8� D 400 lb�T�. Joint D : ∑ Fy D �BD sin ˛�CD � F1 D 0, from which CD D �F1 � BD�0.6� D �300 lb�C� c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 403 Problem 6.20 Consider the truss in Problem 6.19. The loads F1 D 450 lb and F2 D 150 lb. Determine the axial forces in members AB, AC, and BC. Solution: From the solution to Problem 6.19 the angle ˛ D 36.87° and the reaction at E is E D 4F1 C 8F2 6 D 500 lb. Denote the axial force in a member joining two points I, K by IK. Joint E : ∑ Fy D EG D 0. ∑ Fx D AEC E D 0, from which AE D �E D �500 lb�C�. Joint A: ∑ Fx D �AE� AC cos ˛ D 0, from which AC D �AE 0.8 D 625 lb�T� . ∑ Fy D AC sin ˛C AB D 0, from which AB D �AC�0.6� D �375 lb�C� Joint B: ∑ Fy D BD sin ˛� F2 � AB D 0, from which BD D F2 C AB 0.6 D �375 lb�C� ∑ Fx D �BC� BD cos ˛ D 0, from which BC D �BD�0.8� D 300 lb�T� EG E AE AE BC AB BDABAC Joint E Joint A Joint B F2 α α 404 c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.21 Determine the axial forces in members BD, CD, and CE of the truss. C E G FD HA B 4 ft 4 ft 4 ft4 ft4 ft 12 kip Solution: The free-body diagrams for the entire truss as well as for joints A, B and C are shown. From the entire truss: Fx : Ax D 0 FH : �12 kip��8 ft�� Ay�12 ft� D 0 Solving, yields Ax D 0, Ay D 8 kip From joint A: Fx : Ax C TAD cos 45° D 0 Fy : Ay C TAB C TAD sin 45° D 0 Solving yields TAB D �8 kip, TAD D 0 From joint B: Fx : TBD C TBC cos 45° D 0 Fy : TBC C sin 45° � TAB D 0 Solving yields TBD D 8 kip, TBC D �11.3 kip From joint C: Fx : TCE � TBC cos 45° D 0 Fy : �TBC sin 45° � TCD D 0 Solving yields TCD D 8 kip, TCE D �8 kip Thus we have BC : 11.3 kip (C), CD : 8 kip (T), CE : 8 kip (C) c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 405 Problem 6.22 The Warren truss supporting the walkway is designed to support vertical 50-kN loads at B, D, F, and H. If the truss is subjected to these loads, what are the resulting axial forces in members BC, CD, and CE? 6 m6 m6 m6 m A C E G I B D F H 2 m Solution: Assume vertical loads at A and I Find the external loads at A and I, then use the method of joints to work through the structure to the members needed. 3 m 3 m 6 m 6 m 6 m 50 kN 50 kN 50 kN 50 kN x AY IY ∑ Fy : Ay C Iy � 4�50� D 0 (kN) ∑ MA : � 3�50�� 9�50�� 15�50� � 21�50�C 24 Iy D 0 Solving Ay D 100 kN Iy D 100 kN Joint A: y x AB AC A AY θ tan � D 23 � D 33.69° ∑ Fx : AB cos � C AC D 0 ∑ Fy : AB sin � C Ay D 0 Solving, AB D �180.3 kN �C� AC D 150 kN �T� Joint B : 50 kN BD BC AB y B x θ θ AB D �180.3 kN � D 33.69° ∑ Fx : BC cos � C BD � AB cos � D 0 ∑ Fy : � 50� AB sin � � BC sin � D 0 Solving, BC D 90.1 kN �T� BD D �225 kN �C� Joint C : BC AC CE CD y C θ θ x � D 33.69° AC D 150 kN �T� BC D 90.1 kN �T� ∑ Fx : CE� ACCCD cos � � BC cos � D 0 ∑ Fy : CD sin � C BC sin � D 0 Solving, CE D 300 kN �T� CD D �90.1 kN �C� Hence BC D 90.1 kN �T� CD D �90.1 kN �C� CE D 300 kN �T� 406 c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.23 For the Warren truss in Problem 6.22, determine the axial forces in members DF, EF, and FG. Solution: In the solution to Problem 6.22, we solved for the forces in AB, AC, BC, BD, CD, and CE. Let us continue the process. We ended with Joint C. Let us continue with Joint D. Joint D : D BD CD DE θ θ DF x y 50 kN � D 33.69° BD D �225 kN �C� CD D �90.1 kN �C� ∑ Fx : DF� BD C DE cos � � CD cos � D 0 ∑ Fy : � 50�CD sin � � DE sin � D 0 Solving, DF D �300 kN �C� DE D 0 At this point, we have solved half of a symmetric truss with a symmetric load. We could use symmetry to determine the loads in the remaining members. We will continue, and use symmetry as a check. Joint E : CE E EG x y DE EF θ θ � D 33.69° CE D 300 kN �T� DE D 0 ∑ Fx : EG�CEC EF cos � � DE cos � D 0 ∑ Fy : DE sin � C EF sin � D 0 Solving, we get EF D 0 EG D 300 kN �T� Note: The results are symmetric to this point! Joint F : 50 kN EF FG DF F FH x y θ θ � D 33.69° DF D �300 kN �C� EF D 0 ∑ Fx : FH� DFC FG cos � � EF cos � D 0 ∑ Fy : � 50� EF sin � � FG sin � D 0 Solving: FH D �225 kN �C� FG D �90.1 kN �C� Thus, we have DF D �300 kN �C� EF D 0 FG D �90.1 kN �C� Note-symmetry holds! c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 407 Problem 6.24 The Pratt bridge truss supports five forces (F D 300 kN). The dimension L D 8 m. Deter- mine the axial forces in members BC, BI, and BJ. A B C D E G I J K L M H LLL L L L L L F F F F F LL Solution: Find support reactions at A and H. From the free body diagram, ∑ Fx D AX D 0, ∑ Fy D AY CHY � 5�300� D 0, and ∑ MA D 6�8�HY � 300�8C 16C 24C 32C 40� D 0. From these equations, AY D HY D 750 kN. From the geometry, the angle � D 45° Joint A: From the free body diagram, ∑ Fx D AX C TAB cos � C TAI D 0, ∑ Fy D TAB sin � C AY D 0. From these equations, TAB D �1061 kN and TAI D 750 kN. Joint I: From the free body diagram, ∑ Fx D TIJ � TAI D 0, ∑ Fy D TBI � 300 D 0. From these equations, TBI D 300 kN and TIJ D 750 kN. Joint B: From the free body diagram, ∑ Fx D TBC C TBJ cos � � TAB cos � D 0, ∑ Fy D �TBI � TBJ sin � � TAB sin � D 0. From these equations, TBC D �1200 kN and TBJ D 636 kN. B G I J K L M H L L L L L L L F F F F F HYAY L = 8 m F = 300 kN AY A I y x x x y TAI TAB TBI TBC TBJ TBITAB TIJTAI θ θ θ θ F Joint B Joint A Joint I y A 8 8 8 8 8 8 408 c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.25 For the roof truss shown, determine the axial forces in members AD, BD, DE, and DG. Model the supports at A and I as roller supports. A B C F H I E 3 m 3 m 3 m 3 m 3 m 3 m D G 6 kN 6 kN 8 kN 8 kN 10 kN 3.6 m Solution: Use the whole structure to find the reaction at A.∑ MI : �6 kN��3 m�C �8 kN��6 m�C �10 kN��9 m� C �8 kN��12 m�C �6 kN��15 m� C A�18 m� D 0) A D 19 kN 6 kN 8 kN 10 kN 8 kN 6 kN IA Now work with joint A ∑ Fy : FAB sin 21.8° C A D 0) FAB D �51.2 kN ∑ Fx : FAD C FAB cos 21.8° D 0) FAD D 47.5 kN A A FAB FAD 21.8° Next use joint B ∑ Fx : ��FAB C FBC C FBD� cos 21.8° D 0 ∑ Fy : ��FAB C FBC � FBD� sin 21.8° � �6 kN� D 0 Solving: FBC D �43.1 kN, FBD D �8.08 kN 6 kN B FBC FBDFAB Next go to joint C ∑ Fy : ��8 kN�� FCD C �FCE � FBC� sin 21.8° D 0 ∑ Fx : �FCE � FBC� cos 21.8° D 0 Solving: FCD D �8 kN, FCE D �43.1 kN 8 kN C FCD FCDFBC Finally examine joint D ∑ Fx : �FAD C FDG � FBD cos 21.8° C FDE cos 50.19° D 0 ∑ Fy : FBD sin 21.8° C FCD C FDE sin 50.19°D 0 Solving: FDE D 14.3 kN, FDG D 30.8 kN D FCD FDE FDGFAD FBD 50.19° In Summary FAD D 47.5 kN�T�, FBD D 8.08 kN�C�, FDE D 14.32 kN�T�, FDG D 30.8 kN�T� c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 409 Problem 6.26 The Howe truss helps support a roof. Model the supports at A and G as roller supports. Deter- mine the axial forces in members AB, BC, and CD. 4 ft 4 ft 4 ft 4 ft 4 ft 4 ft 800 lb 8 ft A B C G F E D H I J K L 600 lb600 lb 400 lb400 lb Solution: The strategy is to proceed from end A, choosing joints with only one unknown axial force in the x- and/or y-direction, if possible, and if not, establish simultaneous conditions in the unknowns. The interior angles HIB and HJC differ. The pitch angle is ˛Pitch D tan�1 ( 8 12 ) D 33.7°. The length of the vertical members: BH D 4 ( 8 12 ) D 2.6667 ft, from which the angle ˛HIB D tan�1 ( 2.6667 4 ) D 33.7°. CI D 8 8 12 D 5.3333 ft, from which the angle ˛IJC D tan�1 ( 5.333 4 ) D 53.1°. The moment about G: MG D �4C 20��400� C �8C 16��600� C �12��800�� 24A D 0, from which A D 33600 24 D 1400 lb. Check: The total load is 2800 lb. From left-right symmetry each support A, G supports half the total load. check. The method of joints: Denote the axial force in a member joining two points I, K by IK. Joint A: ∑ Fy D AB sin ˛P C 1400 D 0, from which AB D � 1400 sin ˛p D �2523.9 lb �C� ∑ Fx D AB cos ˛Pitch C AH D 0, from which AH D �2523.9��0.8321� D 2100 lb �T� 400 lb 600 lb 800 lb 600 lb 400 lb A G 4 ft 4 ft 4 ft 4 ft 4 ft 4 ft AB AH CD CI CJBC BI HI IJ CI BCBH AH AB BH BI HI1400 lb 400 lb αPitch αPitch αPitch αIJC αPitch αPitch Joint A Joint I Joint C Joint H Joint B 600 lb Joint H : ∑ Fy D BH D 0, or, BH D 0. ∑ Fx D �AHCHI D 0, from which HI D 2100 lb �T� Joint B : ∑ Fx D �AB cos ˛Pitch C BC cos ˛Pitch C BI cos ˛Pitch D 0, from which BCC BI D AB 410 c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 6.26 (Continued ) ∑ Fy D �400� AB sin ˛Pitch C BC sin ˛Pitch � BI sin ˛Pitch D 0, from which BC� BI D ABC 400 sin ˛Pitch . Solve the two simultaneous equations in unknowns BC, BI: BI D � 400 2 sin ˛Pitch D �360.56 lb �C�, and BC D AB� BI D �2163.3 lb �C� Joint I : ∑ Fx D �BI cos ˛Pitch �HIC IJ D 0, from which IJ D 1800 lb �T� ∑ Fy D CBI sin ˛Pitch CCI D 0, from which CI D 200 lb (T) Joint C: ∑ Fx D �BC cos ˛Pitch CCD cos ˛Pitch CCJ cos ˛IJC D 0, from which CD�0.8321� CCJ�0.6� D �1800 ∑ Fy D �600�CI� BC sin ˛Pitch CCD sin ˛Pitch �CJ sin ˛IJC D 0, from which CD�0.5547� �CJ�0.8� D �400 Solve the two simultaneous equations to obtain CJ D �666.67 lb �C�, and CD D �1682.57 lb �C� c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 411 Problem 6.27 The plane truss forms part of the supports of a crane on an offshore oil platform. The crane exerts vertical 75-kN forces on the truss at B, C, and D. You can model the support at A as a pin support and model the support at E as a roller support that can exert a force normal to the dashed line but cannot exert a force parallel to it. The angle ˛ D 45°. Determine the axial forces in the members of the truss. 3.4 m3.4 m 3.4 m3.4 m 1.8 m 2.2 m A E F G H C DB α Solution: The included angles � D tan�1 ( 4 3.4 ) D 49.64°, ˇ D tan�1 ( 2.2 3.4 ) D 32.91°, � D tan�1 ( 1.8 3.4 ) D 27.9°. The complete structure as a free body: The sum of the moments about A is MA D ��75��3.4��1C 2C 3�C �4��3.4�Ey D 0. with this relation and the fact that Ex cos 45° C Ey cos 45° D 0, we obtain Ex D �112.5 kN and Ey D 112.5 kN. From ∑ FAx D Ax C Ex D 0, AX D �EX D 112.5 kN. ∑ FAy D Ay � 3�75�C Ey D 0, from which Ay D 112.5 kN. Thus the reactions at A and E are symmet- rical about the truss center, which suggests that symmetrical truss members have equal axial forces. The method of joints: Denote the axial force in a member joining two points I, K by IK. Joint A: ∑ Fx D AB cos � C Ax C AF cos ˇ D 0, ∑ Fy D AB sin � C Ay C AF sin ˇ D 0, from which two simultaneous equations are obtained. Solve: AF D �44.67 kN �C� , and AB D �115.8 kN �C� Joint E: ∑ Fy D �DE cos � C Ex � EH cos ˇ D 0. ∑ Fy D DE sin � C Ey C EH sin ˇ D 0, from which two simultaneous equations are obtained. AX AX AY EY EX AY EY EX 75 kN 75 kN 75 kN 3.4 m 3.4 m 3.4 m 3.4 m AB AB BF EH AF γ γ γθ θ γβ β ββ DE BF AF FG GH DH EH BG DG DH CG CDBC DE BC 75 kN 75 kN 75 kN CD Joint A Joint E Joint F Joint B Joint D Joint C Joint H Solve: EH D �44.67 kN�C� , and DE D �115.8 kN�C� Joint F : ∑ Fx D �AF cos ˇ C FG D 0, from which FG D �37.5 kN �C� ∑ Fy D �AF sin ˇ C BF D 0, from which BF D �24.26 kN �C� Joint H: ∑ Fx D EH cos ˇ �GH D 0, 412 c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 6.27 (Continued ) from which GH D �37.5 kN �C� ∑ Fy D �EH sin ˇ C DH D 0, from which DH D �24.26 kN �C� Joint B: ∑ Fy D �AB sin � � BFC BG sin � � 75 D 0, from which BG D 80.1 kN �T� ∑ Fx D �AB cos � C BCC BG cos � D 0, from which BC D �145.8 kN �C� Joint D: ∑ Fy D �DE sin � � DH� DG sin � � 75 D 0, from which DG D 80.1 kN �T� ∑ Fx D DE cos � �CD � DG cos � D 0, from which CD D �145.8 kN �C� Joint C : ∑ Fx D CD� BC D 0, from which CD D BC Check. ∑ Fy D �CG� 75 D 0, from which CG D �75 kN �C� Problem 6.28 (a) Design a truss attached to the supports A and B that supports the loads applied at points C and D. (b) Determine the axial forces in the members of the truss you designed in (a) A B C D 2 ft 1000 lb 2000 lb 4 ft 5 ft 5 ft5 ft Problem 6.29 (a) Design a truss attached to the supports A and B that goes over the obstacle and supports the load applied at C. (b) Determine the axial forces in the members of the truss you designed in (a). A B C4 m Obstacle 6 m 3.5 m 4.5 m 1 m 2 m 10 kN Solution: This is a design problem with many possible solutions. c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 413 Problem 6.30 Suppose that you want to design a truss supported at A and B (Fig. a) to support a 3-kN down- ward load at C. The simplest design (Fig. b) subjects member AC to 5-kN tensile force. Redesign the truss so that the largest force is less than 3 kN. A B C A B C 3 kN 1.2 m 1.6 m (a) (b) 3 kN Solution: There are many possible designs. To better understand the problem, let us calculate the support forces in A and B and the forces in the members in Fig. (b). Ax Ay Bx C xB 1.6 m 3 kN 1.2 m A θ tan � D 1.2 1.6 � D 36.87° sin � D 0.6 cos � D 0.8 ∑ Fx: Ax C Bx D 0 ∑ Fy : Ay � 3 kN D 0 C ∑ MA: 1.2Bx � 1.6�3� D 0 Solving, we get Ax D �4 kN Bx D 4 kN Ay D 3 kN Note: These will be the external reactions for every design that we produce (the supports and load do not change). Reference Solution (Fig. (b)) Joint C : θ BC AC 3 kN � D 36.87° ∑ Fx : � BC� AC cos � D 0 ∑ Fy : AC sin � � 3 kN D 0 Solving: BC D �4 kN �C� AC D5 kN �C� Thus, AC is beyond the limit, but BC (in compression) is not, Joint B : BX AB BC ∑ Fx : Bx C BC D 0 ∑ Fy : AB D 0 Solving, BC and Bx are both already known. We get AB D 0 Thus, we need to reduce the load in AC. Consider designs like that shown below where D is inside triangle ABC. Move D around to adjust the load. B C D A However, the simplest solution is to place a second member parallel to AC, reducing the load by half. 414 c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.31 The bridge structure shown in Example 6.2 can be given a higher arch by increasing the 15° angles to 20°. If this is done, what are the axial forces in members AB, BC, CD, and DE? 2b F F F F F bbbb (1) 2b F F F (2) b b b b B C D EA 15�15� G JI KH F F a a Solution: Follow the solution method in Example 6.3. F is known Joint B : α y F x 20° TBC TAB Joint C : F TBC TCD 20°20° C For joint C, ∑ Fx : � TBC cos 20° C TCD cos 20° D 0 ∑ Fy : � F� TBC sin 20° � TCD sin 20° D 0 TBC D TCD D �1.46F �C� For joint B. ∑ Fx : TBC cos 20� TAB cos ˛ D 0 ∑ Fy : TBC sin 20° � F� TAB sin ˛ D 0 Solving, we get ˛ D 47.5° and TAB D �2.03F �C� For the new truss (using symmetry) Members Forces AG, BH, CI, F DJ, EK AB, DE 2.03F (C) BC, CD 1.46F (C) c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 415 Problem 6.32 In Active Example 6.3, use the method of sections to determine the axial forces in members BC, BI and HI. A B C D E F G H I J K L 100 kN M 1 m Solution: The horizontal members of the truss are each 1 m in length. We cut through the relevant members and draw a free-body diagram of the section to the right of the cut. We will use equilibrium equations for this section that are designed to allow us to easily solve for the unknowns. The equilibrium equations MI : TBC�1 m�� �100 kN��4 m� D 0) TBC D 400 kN MB : �THI�1 m�� �100 kN��5 m� D 0) THI D �500 kN Fy : TBI sin 45° � 100 kN D 0) TBI D 141 kN In summary we have BC : 400 kN (T), BI : 141 kN (T), HI : 500 kN (C) Problem 6.33 In Example 6.4, obtain a section of the truss by passing planes through members BE, CE, CG, and DG. Using the fact that the axial forces in members DG and BE have already been determined, use your section to determine the axial forces in members CE and CG. K L L D L L L L G J IC B E H F F2F A Solution: From Example 6.4 we know that TDG D �F, TBE D F Ax D 0, Ay D 2F We make the indicated cuts and isolate the section to the left of the cuts. The equilibrium equations are Fx : TDG C TBE C TCG cos 45° C TCE cos 45° D 0 Fy : Ay � FC TCG sin 45° � TCE sin 45° D 0 Solving yields TCE D Fp 2 , TCG D �Fp 2 We have CE : Fp 2 �T�, CG : Fp 2 �C� 416 c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.34 The truss supports a 100-kN load at J. The horizontal members are each 1 m in length. (a) Use the method of joints to determine the axial force in member DG. (b) Use the method of sections to determine the axial force in member DG. A B C D E F G H 100 kN J 1 m Solution: (a) We draw free-body diagrams of joints J, H, and D. From joint J we have Fy : TDJ sin 45° � �100 kN� D 0 ) TDJ D 141 kN From joint H we have Fy : TDH D 0 From joint D we have Fy : �TDG sin 45° � TDH � TDJ sin 45° D 0 Solving yields TDG D �141 kN (b) We cut through CD, DG and GH. The free-body diagram of the section to the right of the cut is shown. From this diagram we have Fy : �TDG sin 45° � �100 kN� D 0 ) TDG D �141 kN In summary (a), (b) DG : 141 kN (C) c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 417 Problem 6.35 For the truss in Problem 6.34, use the method of sections to determine the axial forces in members BC, CF, and FG. Solution: ∑ Fx: � BC�CF cos 45� FG D 0 ∑ Fy : �CF sin 45° � 100 D 0 ∑ MC: � �1�FG� 2�100� D 0 Solving BC D 300 kN �T� CF D �141.4 kN �C� FG D �200 kN �C� 1 m 45° F FG CF G H1 m 1 m J D BC C 100 kN Problem 6.36 Use the method of sections to determine the axial forces in members AB, BC, and CE. A B C D E G 1 m 1 m 1 m 1 m F 2F Solution: First, determine the forces at the supports AX AY GY B F 2F D C E θ 1 m1 m 1 m 1 m Θ = 45° ∑ Fx: Ax D 0 ∑ Fy : Ay CGy � 3F D 0 C ∑ MA: � 1�F�� 2�2F�C 3Gy D 0 Solving Ax D 0 Gy D 1.67F Ay D 1.33F Method of Sections: AX = 0 AY BC AB CE1 m 1 m y B C F x AY = 1. 33 F AX = 0 ∑ Fx : CEC AB D 0 ∑ Fy : BCC Ay � F D 0 C ∑ MB: ��1�Ay C �1�CE D 0 Solving, we get AB D �1.33F �C� CE D 1.33F �T� BC D �0.33F �C� 418 c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.37 Use the method of sections to determine the axial forces in members DF, EF, and EG. A B C D E F G H 300 mm 400 mm 400 mm 400 mm 400 mm 18 kN 24 kN Solution: We will first use the free-body diagram of the entire structure to find the reaction at F. MB : �18 kN� �400 mm� � �24 kN� �1200 mm� C F �800 mm� D 0 ) F D 27 kN Next we cut through DF, EF, EG and look at the section to the right of the cut. The angle ˛ is given by ˛ D tan�1�3/4� D 36.9° The equilibrium equations are MF : TEG �300 mm�� �24 kN� �400 mm� D 0 ME : �TDF �300 mm�� �24 kN� �800 mm� C F�400 mm� D 0 Fy : F� �24 kN�C TEF sin ˛ D 0 Solving yields TDF D �28 kN, TEF D �5 kN, TEG D 32 kN Thus DF : 28 kN (C), EF : 5 kN (C), EG : 32 kN (T) c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 419 Problem 6.38 The Pratt bridge truss is loaded as shown. Use the method of sections to determine the axial forces in members BD, BE, and CE. 17 ft 17 ft 17 ft 17 ft A B D F H GEC 8 ft 10 kip 30 kip 20 kip Solution: Use the whole structure to find the reaction at A.∑ MH : �20 kip��17 ft�C �30 kip��34 ft� C �10 kip��51 ft�� A�68 ft� D 0 ) A D 27.5 kip Now cut through BD, BE, CE and use the left section ∑ MB : �A�17 ft�C FCE�8 ft� D 0) FCE D 58.4 kip ∑ ME : �10 kip��17 ft�� A�34 ft�� FBD�8 ft� D 0 ) FBD D �95.6 kip ∑ Fy : A� 10 kip� 8p 353 FBE D 0) FBE D 41.1 kip In Summary FCE D 58.4 kip�T�, FBD D 95.6 kip�C�, FBE D 41.1 kip�T� A H10 kip 30 kip 20 kip A C B 8 17 A 10 kip FCE FBE FBD Problem 6.39 The Howe bridge truss is loaded as shown. Use the method of sections to determine the axial forces in members BD, CD, and CE. 17 ft 17 ft 17 ft 17 ft A B D F H GEC 8 ft 10 kip 30 kip 20 kip Solution: Use the whole structure to find the reaction at A (same as 6.38) A D 27.5 kip Now cut through BD, CD, and CE and use the left section. ∑ MC : �A�17 ft�� FBD�8 ft� D 0) FBD D �58.4 kip ∑ MD : �A�34 ft�C �10 kip��17 ft�C FCE�8 ft� D 0 ) FCE D 95.6 kip ∑ Fy : A� 10 kipC 8p 353 FCD D 0) FCD D �41.1 kip In SummaryFBD D 58.4 kip�C�, FCE D 95.6 kip�T�, FCD D 41.1 kip�C� FBD FCD FCE 10 kip A A B C 17 8 420 c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.40 For the Howe bridge truss in Problem 6.39, use the method of sections to determine the axial forces in members DF, DG, and EG. Solution: Same truss as 6.39. Cut through DF, DG, and EG and use left section ∑ MD : �A�34 ft�C �10 kip��17 ft�C FEG�8 ft� D 0 ) FEG D 95.6 kip ∑ MG : �A�51 ft�C �10 kip��34 ft�C �30 kip��17 ft�� FDF�8 ft� D 0) FDF D �69.1 kip ∑ Fy : A� 10 kip� 30 kip� 8p 353 FDG D 0) FDG D �29.4 kip In summary FEG D 95.6 kip�T�, FDF D 69.1 kip�C�, FDG D 29.4 kip�C� FDF D 8 17 FDG FEG E 30 kip10 kipA Problem 6.41 The Pratt bridge truss supports five forces F D 340 kN. The dimension L D 8 m. Use the method of sections to determine the axial force in member JK. A B C D E G I J K L M H LLL L L L L L F F F F F LL Solution: First determine the external support forces. L L L L L L F F F F F AX AY HY F = 340 kN, L = 8 M ∑ Fx : Ax D 0 ∑ Fy : Ay � 5FCHy D 0 C ∑ MA: 6LHy � LF� 2LF� 3LF� 4LF� 5LF D 0 Solving: Ax D 0, Ay D 850 kN Hy D 850 kN Note the symmetry: Method of sections to find axial force in member JK. B A AY L L JI JK K CK D CDC F F θ � D 45° L D 8M F D 340 kN Ay D 850 kN ∑ Fx : CDC JKCCK cos � D 0 ∑ Fy : Ay � 2F�CK sin � D 0 C ∑ MC: L�JK�C L�F�� 2L�Ay� D 0 Solving, JK D 1360 kN �T� Also, CK D 240.4 kN �T� CD D �1530 kN �C� c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 421 Problem 6.42 For the Pratt bridge truss in Prob- lem 6.41, use the method of sections to determine the axial force in member EK. Solution: From the solution to Problem 6.41, the support forces are Ax D 0, Ay D Hy D 850 kN. Method of Sections to find axial force in EK. DE EK E G KL F F HY L θ ∑ Fx : � DE� EK cos � �KL D 0 ∑ Fy : Hy � 2F� EK sin � D 0 ∑ ME: � �L��KL�� �L��F�C �2L�Hy D 0 A B C D E G I J K L M H L L L L L L F F F F F L Solution: EK D 240.4 kN �T� Also, KL D 1360 kN �T� DE D �1530 kN �C� Problem 6.43 The walkway exerts vertical 50-kN loads on the Warren truss at B, D, F, and H. Use the method of sections to determine the axial force in member CE. 6 m6 m6 m6 m A C E G I B D F H 2 m Solution: First, find the external support forces. By symmetry, Ay D Iy D 100 kN (we solved this problem earlier by the method of joints). B BD A y x AY CD D CEC 50 kN 2 m 6 m θ tan � D 2 3 � D 33.69° ∑ Fx: BD CCD cos � CCE D 0 ∑ Fy : Ay � 50CCD sin � D 0 ∑ MC: � 6Ay C 3�50�� 2BD D 0 Solving: CE D 300 kN �T� Also, BD D �225 kN �C� CD D �90.1 kN �C� 422 c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.44 Use the method of sections to determine the axial forces in members AC, BC, and BD. 600 lb D E 3 ft 4 ft 4 ft 3 ft A C B Solution: Obtain a section by passing a plane through members AC, BC, and BD, isolating the part of the truss above the planes. The angle between member AC and the horizontal is ˛ D tan�1�4/3� D 53.3° The equilibrium equations are MC : �600 lb� �4 ft�� TBD cos ˛ �3 ft� D 0 MB : �600 lb� �8 ft�C TAC sin ˛ �4 ft� D 0 Fy : �TBC � TAC cos ˛� TBD cos ˛ D 0 Solving yields TBD D 1000 lb, TAC D �2000 lb, TBC D 800 lb Thus BD : 100 lb (T), AC : 2000 lb (C), BC : 800 lb (T) c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 423 Problem 6.45 Use the method of sections to determine the axial forces in member FH, GH, and GI. I C A B D F H E G 400 mm 400 mm 6 kN 4 kN 400 mm400 mm 300 mm 300 mm Solution: The free-body diagram of the entire truss is used to find the force I. MA : I�600 mm�� �4 kN� �1200 mm� � �6 kN� �800 mm� D 0 ) I D 16 kN Obtain a section by passing a plane through members FH, GH, and GI, isolating the part of the truss to the right of the planes. The angle ˛ is ˛ D tan�1�3/4� D 36.9° The equilibrium equations for the section are MH : TGI cos ˛ �300 mm�C I�300 mm� D 0 MG : I�300 mm�� TFH cos ˛ �400 mm� D 0 Fx : �TGH � TGI sin ˛� TFH sin ˛ D 0 Solving yields TGI D �20 kN, TFH D 20 kN, TGH D �16 kN Thus GI : 20 kN (C), FH : 20 kN (T), GH : 16 kN (C) 424 c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.46 Use the method of sections to determine the axial forces in member DF, DG, and EG. I C A B D F H E G 400 mm 400 mm 6 kN 4 kN 400 mm400 mm 300 mm 300 mm Solution: The free-body diagram of the entire truss is used to find the force I. MA : I�600 mm�� �4 kN� �1200 mm� � �6 kN� �800 mm� D 0 ) I D 16 kN Obtain a section by passing a plane through members DF, DG, and EG, isolating the part of the truss to the right of the planes. The angle ˛ is ˛ D tan�1�3/4� D 36.9° The equilibrium equations for the section are MG : I �300 mm�� TDF�300 mm� D 0 MD : TEG�300 mm�C I�600 mm� � �4 kN��400 mm� D 0 Fy : �TDG sin ˛� �4 kN� D 0 Solving yields TDF D 16 kN, TEG D �26.7 kN, TDG D �6.67 kN Thus DF : 16 kN (T), EG : 26.7 kN (C), DG : 6.67 kN (C) c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 425 Problem 6.47 The Howe truss helps support a roof. Model the supports at A and G as roller supports. (a) Use the method of joints to determine the axial force in member BI. (b) Use the method of sections to determine the axial force in member BI. 2 m 2 m 2 m 2 m 2 m 2 m 2 kN 4 m A B C G F E D H I J K L 2 kN2 kN 2 kN2 kN Solution: The pitch of the roof is ˛ D tan�1 ( 4 6 ) D 33.69°. This is also the value of interior angles HAB and HIB. The complete structure as a free body: The sum of the moments about A is MA D �2�2��1C 2C 3C 4C 5�C 6�2�G D 0, from which G D 30 6 D 5 kN. The sum of the forces: ∑ FY D A� 5�2�CG D 0, from which A D 10� 5 D 5 kN. The method of joints: Denote the axial force in a member joining I, K by IK. (a) Joint A: ∑ Fy D AC AB sin ˛ D 0, from which AB D �A sin ˛ D �5 0.5547 D �9.01 kN (C). ∑ Fx D AB cos ˛C AH D 0, from which AH D �AB cos ˛ D 7.5 kN (T). Joint H : ∑ Fy D BH D 0. Joint B : ∑ Fx D �AB cos ˛C BI cos ˛C BC cos ˛ D 0, ∑ Fy D �2� AB sin ˛� BI sin ˛C BC sin ˛ D 0. Solve: BI D �1.803 kN �C� , BC D �7.195 kN �C� (b) Make the cut through BC, BI and HI. The section as a free body: The sum of the moments about B: MB D �A�2�CHI�2 tan ˛� D 0, from which HI D 3 2 A D 7.5 kN�T�. The sum of the forces: ∑ Fx D BC cos ˛C BI cos ˛CHI D 0, ∑ Fy D A� FC BC sin ˛� BI sin ˛ D 0. Solve: BI D �1.803 kN �C� . F F F F F = 2 kN GA 2 m 2 m 2 m 2 m 2 m 2 m (a) AB A AH HI BI BCF AH HI AB BH α α α Joint A Joint H Joint B BH BI BC 2 kN (b) A 2 m α α αB 426 c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rightsreserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.48 Consider the truss in Problem 6.47. Use the method of sections to determine the axial force in member EJ. Solution: From the solution to Problem 6.47, the pitch angle is ˛ D 36.69°, and the reaction G D 5 kN. The length of member EK is LEK D 4 tan ˛ D 166 D 2.6667 m. The interior angle KJE is ˇ D tan�1 ( LEK 2 ) D 53.13°. Make the cut through ED, EJ, and JK. Denote the axial force in a member joining I, K by IK. The section as a free body: The sum of the moments about E is ME D C4G� 2�F�� JK�2.6667� D 0, from which JK D 20� 4 2.6667 D 6 kN �T�. The sum of the forces: ∑ Fx D �DE cos ˛� EJ cos ˇ � JK D 0. ∑ Fy D DE sin ˛� EJ sin ˇ � 2FCG D 0, from which the two simultaneous equations: 0.8321DE C 0.6EJ D �6, 0.5547DE � 0.8EJ D �1. Solve: EJ D �2.5 kN �C� . DE F E β α EJ JK F G 2 m 2 m c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 427 Problem 6.49 Use the method of sections to determine the axial forces in member CE, DE, and DF. C E G FD HA B 4 ft 4 ft 4 ft4 ft4 ft 12 kip Solution: The free-body diagrams for the entire structure and the section to the right of the cut are shown. From the entire structure: MA : ��12 kip� �4 ft� H �12 ft� D 0 ) H D 4 kip Using the section to the right of the cut we have ME : H�4 ft�� TDF�4 ft� D 0 MD : H�8 ft�C TCE�4 ft� D 0 Fy : H� TDE sin 45° D 0 Solving yields TDF D 4 kip, TCE D �8 kip, TDE D 5.66 kip Thus we have DF : 4 kip (T) CE : 8 kip (C) DE : 5.66 kip (T) 428 c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.50 For the bridge truss shown, use the method of sections to determine the axial forces in members CE, CF, and DF. D F H J I 200 kN 200 kN 200 kN 200 kN 200 kN B A C E G 3 m 4 m 7 m 5 m 5 m 5 m 5 m Solution: From the entire structure we find the reactions at A∑ Fx : Ax D 0 ∑ MI : �200 kN��5 m�C �200 kN��10 m�C �200 kN��15 m� C �200 kN��20 m�� Ay�20 m� D 0) Ay D 500 kN 200 kN I 200 kN 200 kN 200 kN 200 kN Ax Ay Now we cut through DF, CF, and CE and use the left section. ∑ MC : �200 kN��5 m�� Ay�5 m�C Ax�3 m�� FDF�4 m� D 0 ) FDF D �375 kN ∑ MF : �200 kN��10 m�C �200 kN��5 m�� Ay�10 m�C Ax�7 m� C 5p 26 FCE�4 m�� 1p 26 FCE�5 m� D 0) FCE D 680 kN ∑ Fx : Ax C FDF C 5p 26 FCE C 5p 41 FCF D 0 ) FCF D �374 kN FDF FCF Ay Ax 4 5 5 1 FCE 200 kN 200 kN D C Summary: FDF D 375 kN�C�, FCE D 680 kN�T�, FCF D 374 kN�C� c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 429 Problem 6.51 The load F D 20 kN and the dimension L D 2 m. Use the method of sections to determine the axial force in member HK. Strategy: Obtain a section by cutting members HK, HI, IJ, and JM. You can determine the axial forces in members HK and JM even though the resulting free- body diagram is statically indeterminate. A B C D H K G J M E I F F L L L L L Solution: The complete structure as a free body: The sum of the moments about K is MK D �FL�2C 3�CML�2� D 0, from which M D 5F 2 D 50 kN. The sum of forces: ∑ FY D KY CM D 0, from which KY D �M D �50 kN.∑ FX D KX C 2F D 0, from which KX D �2F D �40 kN. The section as a free body: Denote the axial force in a member joining I, K by IK. The sum of the forces: ∑ Fx D Kx �HIC IJ D 0, from which HI� IJ D Kx . Sum moments about K to get MK D M�L��2�C JM�L��2�� IJ�L�CHI�L� D 0. Substitute HI� IJ D Kx , to obtain JM D �M� Kx 2 D �30 kN �C�. ∑ Fy D Ky CMC JMCHK D 0, from which HK D �JM D 30 kN�T� F 2L 2L 2L F MKX KX KY KY L HI IJ HK JM M L 430 c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.52 The weight of the bucket is W D 1000 lb. The cable passes over pulleys at A and D. (a) Determine the axial forces in member FG and HI. (b) By drawing free-body diagrams of sections, explain why the axial forces in members FG and HI are equal. 3 ft 6 in 3 ft 3 ft 3 ft 3 in 35° L J H F C K I G E B AD W Solution: The truss is at angle ˛ D 35° relative to the horizontal. The angles of the members FG and HI relative to the horizontal are ˇ D 45° C 35° D 80°. (a) Make the cut through FH, FG, and EG, and consider the upper section. Denote the axial force in a member joining, ˛, ˇ by ˛ˇ. The section as a free body: The perpendicular distance from point F is LFW D 3 p 2 sin ˇ C 3.5 D 7.678 ft. The sum of the moments about F is MF D �WLFW CW�3.25�� jEGj�3� D 0, from which EG D �1476.1 lb �C�. The sum of the forces: ∑ FY D �FG sin ˇ � FH sin ˛� EG sin ˛�W sin ˛�W D 0, ∑ FX D �FG cos ˇ � FH cos ˛� EG cos ˛�W cos ˛ D 0, from which the two simultaneous equations: �0.9848FG � 0.5736FH D 726.9, and �0.1736FG � 0.8192FH D �389.97. Solve: FG D �1158.5 lb �C� , and FH D 721.64 lb �T�. Make the cut through JH, HI, and GI, and consider the upper section. The section as a free body: The perpendicular distance from point H to the line of action of the weight is LHW D 3 cos ˛C 3 p 2 sin ˇ C 3.5 D 10.135 ft. The sum of the moments about H is MH D �W�L�� jGIj�3�CW�3.25� D 0, from which jGIj D �2295 lb �C�. ∑ FY D �HI sin ˇ � JH sin ˛�GI sin ˛�W sin ˛�W D 0, ∑ FX D �HI cos ˇ � JH cos ˛�GI cos ˛�W cos ˛ D 0, from which the two simultaneous equations: �0.9848HI� 0.5736JH D 257.22, and �0.1736HI� 0.8192JH D �1060.8. Solve: HI D �1158.5 lb�C� , and JH D 1540.6 lb�T� . W W W W FH FG JH HI GI EG α β 3.25 ft 3 ft 3.5 ft (b) Choose a coordinate system with the y axis parallel to JH. Isolate a section by making cuts through FH, FG, and EG, and through HJ, HI, and GI. The free section of the truss is shown. The sum of the forces in the x- and y-direction are each zero; since the only external x-components of axial force are those contributed by FG and HI, the two axial forces must be equal: ∑ Fx D HI cos 45° � FG cos 45° D 0, from which HI D FG c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 431 Problem 6.53 Consider the truss in Problem 6.52. The weight of the bucket is W D 1000 lb. The cable passes over pulleys at A and D. Determine the axial forces in members IK and JL. Solution: Make a cut through JL, JK, and IK, and consider the upper section. Denote the axial force in a member joining, ˛, ˇ by ˛ˇ. The section as a free body: The perpendicular distance from point J to the line of action of the weight is L D 6 cos ˛C 3p2 sin ˇ C 3.5 D 12.593 ft. The sum of the moments about J is MJ D �W�L�C W�3.25�� IK�3� D 0, from which IK D �3114.4 lb�C�. The sum of the forces: ∑ Fx D JL cos ˛� IK cos ˛ �W cos ˛� JK cos ˇ D 0, and ∑ Fy D �JL sin ˛� IK sin ˛ �W sin ˛�W� JK sin ˇ D 0, from which two simultaneous equations: 0.8192JL C 0.1736JK D �1732 and 0.5736JL C 0.9848JK D 212.75. Solve: JL D 2360 lb�T� , and JK D �1158.5lb�C� . W W 3.5 ft 3 ft 3.25 ft β αJL JK IK Problem 6.54 The truss supports loads at N, P, and R. Determine the axial forces in members IL and KM. 2 m 2 m 2 m 2 m 1 m 6 m 2 m 2 m 2 m 2 m 2 m K I M L O N Q P RJ H F D G E C BA 1 kN 2 kN 1 kNSolution: The strategy is to make a cut through KM, IM, and IL, and consider only the outer section. Denote the axial force in a member joining, ˛, ˇ by ˛ˇ. The section as a free body: The moment about M is MM D �IL � 2�1�� 4�2�� 6�1� D 0, from which IL D �16 kN �C� . The angle of member IM is ˛ D tan�1�0.5� D 26.57°. The sums of the forces: ∑ Fy D �IM sin ˛� 4 D 0, from which IM D � 4 sin ˛ D �8.944 kN (C). ∑ Fx D �KM� IM cos ˛� IL D 0, from which KM D 24 kN�T� α KM IM IL 1 kN 2 kN 1 kN 1 m 2 m 2 m 2 m 432 c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.55 Consider the truss in Problem 6.54. Determine the axial forces in members HJ and GI. Solution: The strategy is to make a cut through the four members AJ, HJ, HI, and GI, and consider the upper section. The axial force in AJ can be found by taking the moment of the structure about B. The complete structure as a free body: The angle formed by AJ with the vertical is ˛ D tan�1 ( 4 8 ) D 26.57° . The moment about B is MB D 6AJ cos ˛� 24 D 0, from which AJ D 4.47 kN (T). The section as a free body: The angles of members HJ and HI relative to the vertical are ˇ D tan�1 ( 2 8 ) D 14.0°, and � D tan�1 ( 1.5 2 ) D 36.87° respectively. Make a cut through the four members AJ, HJ, HI, and GI, and consider the upper section. The moment about the point I is MI D �24C 2AJ cos ˛C 2HJ cos ˇ D 0. From which HJ D 8.25 kN �T� . The sums of the forces: ∑ Fx D �AJ sin ˛CHJ sin ˇ�HI sin � D 0, from which HI D AJ sin ˛�HJ sin ˇ sin � D 2� 2 sin � D 0. ∑ FY D �AJ cos ˛�HJ cos ˇ �HI cos � �GI� 4 D 0, from which GI D �16 kN �C� AJ HJ HI GI 2 m 2 m 2 m 1 kN 2 kN 1 kN 2 m 2 m 1 m I γα β Problem 6.56 Consider the truss in Problem 6.54. By drawing free-body diagrams of sections, explain why the axial forces in members DE, FG, and HI are zero. Solution: Define ˛, ˇ to be the interior angles BAJ and ABJ respectively. The sum of the forces in the x-direction at the base yields AX C BX D 0, from which Ax D �Bx . Make a cut through AJ, BD and BC, from which the sum of forces in the x-direction, Ax � BD sin ˇ D 0. Since Ax D AJ sin ˛, then AJ sin ˛� BD sin ˇ D 0. A repeat of the solution to Problem 6.55 shows that this result holds for each section, where BD is to be replaced by the member parallel to BD. For example: make a cut through AJ, FD, DE, and CE. Eliminate the axial force in member AJ as an unknown by taking the moment about A. Repeat the solution process in Problem 6.55, obtaining the result that DE D AJ sin ˛� DF sin ˇ cos �DE D 0 where �DE is the angle of the member DE with the vertical. Similarly, a cut through AJ, FH, FG, and EG leads to FG D AJ sin ˛� FH sin ˇ cos �FG D 0, and so on. Thus the explanation is that each member BD, DF, FH and HJ has equal tension, and that this tension balances the x-component in member AJ c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 433 Problem 6.57 In Active Example 6.5, draw the free- body diagram of joint B of the space truss and use it to determine the axial forces in members AB, BC, and BD. 1200 lb B D (10, 0, 0) ft C (6, 0, 6) ft A (5, 3, 2) ft z y x Solution: From Active Example 6.5 we know that the vertical reaction force at B is 440 lb. The free-body diagram of joint B is shown. We have the following position vectors. rBA D �5iC 3jC 2k� ft rBC D �6iC 6k� ft rBD D �10i� ft The axial forces in the rods can then be written as TAB rBA jrBAj D TAB�0.811iC 0.487jC 0.324k� TBC rBC jrBCj D TBC�0.707iC 0.707k� TBD rBD jrBDj D TBDi The components of the equilibrium equations are Fx : 0.811TAB C 0.707TBC C TBD D 0 Fy : 0.487TAB C 440 lb D 0 Fz : 0.324TAB C 0.707TBC D 0 Solving yields TAB D �904 lb, TBC D 415 lb, TBD D 440 lb Thus AB : 904 lb (C), BC : 415 lb (T), BD : 440 lb (T) 434 c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.58 The space truss supports a vertical 10- kN load at D. The reactions at the supports at joints A, B, and C are shown. What are the axial forces in the members AD, BD, and CD? B (5, 0, 3) m AyAx Az Cy Cz By C (6, 0, 0) m D (4, 3, 1) m 10 kN z y x A Solution: Consider the joint D only. The position vectors parallel to the members from D are rDA D �4i� 3j� k, rDB D i� 3jC 2k, rDC D 2i� 3j� k. The unit vectors parallel to the members from D are: eDA D rDAjrDAj D �0.7845i� 0.5883j � 0.1961k eDB D rDBjrDBj D 0.2673i� 0.8018j C 0.5345k eDC D rDCjrDCj D 0.5345i� 0.8018j � 0.2673k The equilibrium conditions for the joint D are ∑ F D TDAeDA C TDBeDB C TDCeDC � FD D 0, from which ∑ Fx D �0.7845TDA C 0.2673TDB C 0.5345TDC D 0 ∑ Fy D �0.5883TDA � 0.8018TDB � 0.8108TDC � 10 D 0 ∑ Fz D �0.1961TDA C 0.5345TDB � 0.2673TDC D 0. Solve: TDA D �4.721 kN �C� , TDB D �4.157 kN �C� TDC D �4.850 kN �C� 10 kN TDC TDB TDA c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 435 Problem 6.59 Consider the space truss in Prob- lem 6.58. The reactions at the supports at joints A, B, and C are shown. What are the axial forces in members AB, AC, and AD? Solution: The reactions at A are required for a determination of the equilibrium conditions at A. The complete structure as a free body: The position vectors are rAB D 5iC 3k, rAC D 6i, rAD D 4iC 3jC k. The sum of the forces: ∑ Fx D Ax D 0, ∑ Fy D Ay C Cy C By � 10 D 0, and ∑ Fz D Az CCz D 0. The moments due to the reactions: M D rAB ð FB C rAC ð FC C rAD ð FD D 0 M D ∣∣∣∣∣∣ i j k 5 0 3 0 By 0 ∣∣∣∣∣∣C ∣∣∣∣∣∣ i j k 6 0 0 0 Cy Cz ∣∣∣∣∣∣C ∣∣∣∣∣∣ i j k 4 3 1 0 �10 0 ∣∣∣∣∣∣ D 0 D ��3By C 10�i� �6Cz�jC �5By C 6Cy � 40�k D 0. These equations for the forces and moments are to be solved for the unknown reactions. The solution: Ax D Cz D 0, Ay D 2.778 kN, By D 3.333 kN, and Cy D 3.889 kN The method of joints: Joint A: The position vectors are given above. The unit vectors are: eAB D 0.8575iC 0.5145k, eAC D i, eAD D 0.7845iC 0.5883jC 0.1961k. The equilibrium conditions are: ∑ F D TABeAB C TAC C eAC C TADeAD C A D 0, from which ∑ Fx D 0.8575TAB C TAC C 0.7845TAD D 0 ∑ Fy D 0TAB C 0TAC C 0.5883TAD C 2.778 D 0 ∑ Fz D 0.5145jTABj C 0jTACj C 0.1961jTADj D 0. Solve: TAB D 1.8 kN �T� , TAC D 2.16 kN �T� TAD D �4.72 kN �C� Ay Ax Az TAB TAC TAD 436 c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.60 The space truss supports a vertical load F at A. Each member is of length L, and the truss rests on the horizontal surface on roller supports at B, C, and D. Determine the axial forces in members AB, AC, and AD. F A B C D Solution: By symmetry, the axial forces in members AB, AC, and AD are equal. We just needto determine the angle � between each of these members and the vertical: F A TAB TAC = TAB TAD = TAB θ θ θ FC 3TAB cos � D 0, so TAB D TAC D TAD D � F3 cos � . From the top view, L C b 60° 30° L /2 we see that b( L 2 ) D tan 30° and bC c( L 2 ) D tan 60°, from which we obtain c D 1 2 L�tan 60° � tan 30°�. Then � D arcsin ( c L ) D 35.26° and TAB D TAC D TAD D � F 3 cos 35.26° D �0.408F. c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 437 Problem 6.61 For the truss in Problem 6.60, deter- mine the axial forces in members AB, BC, and BD. Solution: See the solution of Problem 6.60. The axial force in member AB is TAB D �0.408F, and the angle between AB and the vertical is � D 35.26°. The free-body diagram of joint B is TAB TBC TBD = TBC θ 30° 30° From the equilibrium equation TAB sin � C 2TBC cos 30° D 0, we obtain TBC D TBD D 0.136F. Problem 6.62 The space truss has roller supports at B, C, and D and supports a vertical 800-lb load at A. What are the axial forces in members AB, AC, and AD? 800 lb B D (6, 0, 0) ft C (5, 0, 6) ft A (4, 3, 4) ft z y x Solution: The position vectors of the points A, B, C, and D are rA D 4iC 3jC 4k, rC D 5iC 6k, rD D 6i. The position vectors from joint A to the vertices are: rAB D rB � rA D �4i� 3j� 4k, rAC D rC � rA D 1i� 3jC 2k, rAD D rD � rA D 2i� 3j� 4k Joint A: The unit vectors parallel to members AB, AC, and AD are eAB D rABjrABj D �0.6247i� 0.4685j� 0.6247k, eAC D rACjrACj D 0.2673i� 0.8018j C 0.5345k, and eAD D rADjrADj D 0.3714i� 0.5570j � 0.7428k. The equilibrium conditions at point A: ∑ Fx D �0.6247TAB C 0.2673TAC C 0.3714TAD D 0 ∑ Fy D �0.4685TAB � 0.8018TAB � 0.5570TAD � 800 D 0 ∑ Fz D �0.6247TAB C 0.5345TAC � 0.7428TAD D 0. 800 lb TAD TAC TAB Solve: TAB D �379.4 lb �C� , TAC D �665.2 lb �C� , and TAD D �159.6 lb �C� 438 c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.63 The space truss shown models an airplane’s landing gear. It has ball and socket supports at C, D, and E. If the force exerted at A by the wheel is F D 40j (kN), what are the axial forces in members AB, AC, and AD? B (1, 0, 0) m A (1.1, –0.4, 0) m 0.4 m 0.6 m y x z E (0, 0.8, 0) m C D F Solution: The important points in this problem are A (1.1, �0.4, 0), B (1, 0, 0), C (0, 0, 6), and D (0, 0, �0.4). We do not need point E as all of the needed unknowns converge at A and none involve the location of point E. The unit vectors along AB, AC, and AD are uAB D �0.243iC 0.970jC 0k, uAC D �0.836iC 0.304jC 0.456k, and uAD D �0.889iC 0.323j� 0.323k. The forces can be written as TRS D TRSuRS D TRSXiC TRSYjC TRSZk, where RS takes on the values AB, AC, and AD. We now have three forces written in terms of unknown magnitudes and known directions. The equations of equilibrium for point A are∑ Fx D TABuABX C TACuACX C TADuADX C FX D 0, ∑ Fy D TABuABY C TACuACY C TADuADY C FY D 0, and ∑ Fz D TABuABZ C TACuACZ C TADuADZ C FZ D 0, where F D FXiC FYjC FZk D 40j kN. Solving these equations for the three unknowns, we obtain TAB D �45.4 kN (compression), TAC D 5.26 kN (tension), and TAD D 7.42 kN (tension). y z x E D C B F A (0, 0.8, 0) m 0.4 m 0.6 m (1, 0, 0) m (1.1, −0.4, 0) m TABTAD TAC c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 439 Problem 6.64 If the force exerted at point A of the truss in Problem 6.63 is F D 10iC 60jC 20k (kN), what are the axial forces in members BC, BD and BE? Solution: The important points in this problem are A (1.1, �0.4, 0), B (1, 0, 0), C (0, 0, 0.6), D (0, 0, �0.4), and E (0, 0.8, 0). The unit vectors along AB, AC, AD, BC, BD, and BE are uAB D �0.243iC 0.970jC 0k, uAC D �0.836iC 0.304jC 0.456k, uAD D �0.889iC 0.323j� 0.323k, uBC D �0.857iC 0jC 0.514k, uBD D �0.928iC 0j� 0.371k, and uBE D �0.781iC 0.625jC 0k. The forces can be written as TRS D TRSuRS D TRSXiC TRSYjC TRSZk, where RS takes on the values AB, AC, and AD when dealing with joint A and AB, BC, BD, and BD when dealing with joint B. We now have three forces written in terms of unknown magnitudes and known directions. Joint A: The equations of equilibrium for point A are, ∑ Fx D TABuABX C TACuACX C TADuADX C FX D 0, ∑ Fy D TABuABY C TACuACY C TADuADY C FY D 0, and ∑ Fz D TABuABZ C TACuACZ C TADuADZ C FZ D 0, where F D FXiC FYjC FZk D 10iC 60jC 20k kN. Solving these equations for the three unknowns at A, we obtain TAB D �72.2 kN (compression), TAC D �13.2 kN (compression), and TAD D 43.3 kN (tension). Joint B: The equations of equilibrium at B are ∑ Fx D �TABuABX C TBCuBCX C TBDuBDX C TBEuBEX D 0, ∑ Fy D �TABuABY C TBCuBCY C TBDuBDY C TBEuBEY D 0, and ∑ Fz D �TABuABZ C TBCuBCZ C TBDuBDZ C TBEuBEZ D 0. Since we know the axial force in AB, we have three equations in the three axial forces in BC, BD, and BE. Solving these, we get TBC D 32.7 kN (tension), TBD D 45.2 kN (tension), and TBE D �112.1 kN (compression). y z x E C D B F A (0, 0.8, 0) m 0.4 m 0.6 m (1, 0, 0) m (1.1, −0.4, 0) m TAB TDETAD TBC 440 c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.65 The space truss is supported by roller supports on the horizontal surface at C and D and a ball and socket support at E. The y axis points upward. The mass of the suspended object is 120 kg. The coordinates of the joints of the truss are A: (1.6, 0.4, 0) m, B: (1.0, 1.0, �0.2) m, C: (0.9, 0, 0.9) m, D: (0.9, 0, �0.6) m, and E: (0, 0.8, 0) m. Determine the axial forces in members AB, AC, and AD. x y z B A D C E Solution: The important points in this problem are A: (1.6, 0.4, 0) m, B: (1, 1, �0.2) m, C: (0.9, 0, 0.9) m, and D: (0.9, 0, �0.6) m. We do not need point E as all of the needed unknowns converge at A and none involve the location of point E. The unit vectors along AB, AC, and AD are uAB D �0.688iC 0.688j� 0.229k, uAC D �0.579i� 0.331jC 0.745k, and uAD D �0.697i� 0.398j� 0.597k. The forces can be written as TRS D TRSuRS D TRSXiC TRSYjC TRSZk, where RS takes on the values AB, AC, and AD. We now have three forces written in terms of unknown magnitudes and known directions. The equations of equilibrium for point A are ∑ Fx D TABuABX C TACuACX C TADuADX C FX D 0, ∑ Fy D TABuABY C TACuACY C TADuADY C FY D 0, and ∑ Fz D TABuABZ C TACuACZ C TADuADZ C FZ D 0, where F D FXiC FYjC FZk D �mgj D �1177j N. Solving these equations for the three unknowns, we obtain TAB D 1088 N (tension), TAC D �316 N (compression), and TAD D �813 N (compression). y x E B D C z A mg TAB TAD TAC L c� 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 441 Problem 6.66 The free-body diagram of the part of the construction crane to the left of the plane is shown. The coordinates (in meters) of the joints A, B, and C are (1.5, 1.5, 0), (0, 0, 1), and (0, 0, �1), respectively. The axial forces P1, P2, and P3 are parallel to the x axis. The axial forces P4,
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