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6.1 Problem Determine the axial forces in the members of the truss and indicate whether they are in tension (T) or compression (C). Strategy: Draw free-body diagram of joint A. By writing the equilibrium equations for the joint, you can determine the axial forces in the two members. A 800 N B 30 60 C Solution: A 800 N y 800 N x 30 FAB A 60 B 30 60 C Solving: We get FAC FAB = 693 N (tension) FAC = -400 N (compression) Assume the forces are in the directions shown (both in tension). If a force turns up negative, that force will be in compression. Equilibrium Eqns. Fx : Fy : -FAB cos 30 + FAC cos 60 + 800 = 0 -FAB sin 30 - FAC sin 60 = 0 Problem 6.2 The truss supports a 10-kN load at C. (a) Draw the free-body diagram of the entire truss, and determine the reactions at its supports. (b) Determine the axial forces in the members. Indicate whether they are in tension (T) or compression (C). B 3m A C 4m 10 kN Solution: (a) The free-body diagram of the system is shown. The sum of the moments about B is: MB = 3Ax - 4(10) = 0, from which Ax = 13.33 kN. The sums of the forces: Fx = Ax + Bx = 0, from which Bx = -Ax = -13.33 kN . Fy = By - 10 = 0, from which By = 10 kN . (b) The interior angle ACB is = tan-1 (0.75) = 36.87 . (b) Assume that the unknown forces act away from the joint. Denote the axial force in the member I, K by IK. The axial forces are FCB = BC(-i cos + j sin ), and FCA = -ACi. Summing the forces: Fy = BC sin - 10 = 0, from which BC = 16.67 kN (T ) . Fx = -BC cos - AC = 0, from which AC = -13.33 kN (C) . For the joint A, BX 3m 3m B A 4m C 10 kN BY AX 4m W Fy = AB = 0, from which AB = 0 Problem 6.3 In Example 6.1, suppose that the 2-kN load is applied at D in the horizontal direction, pointing from D toward B. What are the axial forces in the members? Solution: First, solve for the support forces and then use the method of joints at each joint to solve for the forces. AY AX 6m 10 m BX B Fx : Fy : MB : Bx + Ax - 2 kN = 0 Ay = 0 -6Ax = 0 A 3m C 3m D 2 kN y tan = 6 10 = 30.96 D 2 kN x B 0 5m 5m Fy : AC /sin - BC sin - CD sin = 0 -BC - CD = 0 Solving, we get Ax = Ay = 0, Bx = 2 kN Joint A: Solving, we get BC = CD = 0 Joint B: AB BC AY = 0 = 30.96 BX BD x We already know AB = BC = 0 and Bx = 2 kN Fx : BC /cos + BD + Bx = 0 BD + 2 kN = 0 BD = -2 kN (compression) Fy : (all forces zero) = 0 0 AX = 0 AB AC = 30.96 Fx : Fy : AC cos = 0 -AB - AC sin = 0 we have AB = AC = BC = CD = 0 BD = -2 kN (c) Solving, we get AB = AC = 0 Joint C: Note that we did not have to use joint D as we had already solved for the forces there. The F BD at D is BD, with the (-) sign, is opposite the direction shown. AC AC = 0 from above = 30.96 C CD = 0 D 2 kN BD = -2 kN (c) CD BC 0 Fx : AC /cos - BC cos + CD cos = 0 -BC + CD = 0 Problem 6.4 Determine the axial forces in the members of the truss. 0.3 m B A 2 kN 0.4 m C 0.6 m 1.2 m Solution: First, solve for the support reactions at B and C, and then use the method of joints to solve for the forces in the members. BY 0.3 m BX 0.4 m 0.6 m 0.6 m CY Fx : Fy : + MB : Bx + 2 kN = 0 B y + Cy = 0 0, 6Cy - (0, 3)(2 kN) = 0 A 0.3 m B 2 kN A 2 kN 0.4 m C 1.2 m 1.2 m tan = 7 12 (BC = -0.961 kN) = 30.26 Fx = -BC cos + AC cos = 0 Fy = BC sin + AC sin + 1 = 0 Solving, we get AC = -0.926 kN We have AB = 2.839 kN (T ) BC = -0.961 kN (C) Solving, Bx = -2 kN Cy = 1 kN By = -1 kN Joint B: BY (-1 kN) y BX (-2 kN) AB 6 3 AC = -0.926 kN (C) Check: Look at Joint A 18 4 x y 2 kN x AB BC Fx : Fy : AB cos + BC cos - 2 = 0 AB sin - BC sin - 1 = 0 Solving, we get AB = 2.839 kN, BC = -0.961 kN Joint C: AC AC BC 7 Fx : Fy : -AB cos - AC cos + 2 = 0 -AB sin - AC sin = 0 12 1 kN Substituting in the known values, the equations are satisfied: Check! Problem 6.5 (a) Let the dimension h = 0.1 m. Determine the axial forces in the members, and show that in this case this truss is equivalent to the one in Problem 6.4. (b) Let the dimension h = 0.5 m. Determine the axial forces in the members. Compare the results to (a), and observe the dramatic effect of this simple change in design on the maximum tensile and compressive forces to which the members are subjected. B A h D 1 kN 0.4 m C 0.6 m 1.2 m 0.7 m Solution: To get the force components we use equations of the form TP Q = TP Q eP Q = TP QX i + TP QY j where P and Q take on the designations A, B, C, and D as needed. Equilibrium yields At joint A: Fx = TABX + TACX = 0, and Fy = TABY + TACY - 1 kN = 0. Fx = -TABX + TBCX + TBDX = 0, and Fy = -TABY + TBCY + TBDY = 0. Fx = -TBCX - TACX + TCDX = 0, and Fy = -TBCY - TACY + TCDY + CY = 0. Fx = -TCDX - TBDX + DX = 0, and Fy = -TCDY - TBDY + DY = 0. y h DX D 0.4 m DY B A 1 kN C 0.6 m CY 1.2 m 0.7 m x At joint B: At joint C: y B -TAB TAB h 0.4 m At joint D: -TBD TBD T BC TAC DX D -TBC T -TCD CD -TAC DY C CY 0.6 m 1.2 m A 1 kN 0.7 m x Solve simultaneously to get TAB = TBD = 2.43 kN, TAC = -2.78 kN, TBC = 0, TCD = -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. (b) For this part of the problem, we set h = 0.5 m. The unit vectors change because h is involved in the coordinates of point B. The new unit vectors are eAB = -0.986i + 0.164j, eAC = -0.864i - 0.504j, eBC = 0i - 1j, eBD = -0.768i - 0.640j, and eCD = -0.832i + 0.555j. We get the force components as above, and the equilibrium forces at the joints remain the same. Solving the equilibrium equations simultaneously for this situation yields TAB = 1.35 kN, TAC = -1.54 kN, TBC = -1.33, TBD = 1.74 kN, and TCD = -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. Problem 6.6 The load F = 10 kN. Determine the axial forces in the members. B 3m A C D F 4m 4m Solution: TBD The free-body diagram of joint D is From the equations Fx = -TAB cos + TBD cos = 0, Fy = -TBC - TAB sin - TBD sin = 0, TCD F where = arctan(3/4) = 36.9 . From the equations Fx = -TCD - TBD cos = 0, Fy = TBD sin - F = 0, we obtain TBD = 1.67F = 16.7 kN, TCD = -1.33F = -13.3 kN. Joint B we obtain TAB = 1.67F = 16.7 kN, TBC = -2F = -20 kN. Joint C TBC TAC TCD C we see that TAB TBD TBC TAC = TCD = -1.33F = -13.3 kN. Problem 6.7 Consider the truss in Problem 6.6. Each member will safely support a tensile force of 150 kN and a compressive force of 30 kN. What is the largest downward load F that the truss will safely support at D? Solution: See the solution of Problem 6.6. The largest tensile load is 1.67F in members BD and AB. Setting 1.67F = 150 kN gives F = 90 kN. The largest compressive load is 2F in member BC. Setting 2F = 30 kN gives F = 15 kN. The largest load is F = 15 kN. Problem 6.8 The Howe and Pratt bridge trusses are subjected to identical loads. (a) In which truss does the largest tensile force occur? In what member(s) does it occur, and what is its value? (b) In which truss does the largest compressive force occur? In what member(s) does it occur, and what is its value? L B L C L D L L A G F H F Howe I F E Solution: (a) Howe Bridge: The moment about A is MA = -6F + 4E = 0, from which E = 3 F . Denote the axial force in the 2 member I, K by IK. E (1) Joint E: DE = - sin 45 = - 3 2 2 F (C) EI = DE cos 45 = (2) 3 F (T ) 2 3 F (T ) 2 =- L B L C L D L L A G F 2 F (C), 2 E H F Pratt L A B G F L C H F L D I F L L E I F Joint D: CD = DE cos 45 = - 3 F (C), 2 DI = -DE sin 45 = (3) Joint I: CI = F -|DI| sin 45 HI = EI - CI sin 45 = 2F (T ) (4) Joint H: CH = F (T ), GH = HI = 2F (T ). By symmetry (the reaction at A has no x-component) the axial forces in the other members are HG = HI, CG = CI, BG = DI, CD = BC, AG = EI, and AB = DE. In the Howe truss, the members HI and GH have the highest tensile force GH = HI = 2F (T ) and the members DE and AB have the highest compressive force AB = DE = - 3 2 2 F (C) (b) Pratt Bridge: The moment about A is MA = -6F +4E = 0, from which E = 3 F . 2 (1) (2) (3) Joint E: DE = - 3 2 2 F (C), EI = Joint I: DI = F (T ), HI = Joint D: DH = - 2DI - DH 2 -DE 2 L B L A G F L C L D L H F I E F L L L L = 3 F (T ) 2 L EI = 3 F (T ) 2 DE = 22 F (T ), DC = = -2F (C) DE 2 - AX AY Howe Bridge DE 45 F F F E (4) (5) Joint C: BC = DC = -2F (C), CH = 0 Joint H: GH = HI = 3 F (T ). The axial forces in the remain2 ing members are determined from symmetry. In the Pratt Bridge, the highest tensile force occurs in members EI, HI, GH, and AG = 3 F (T ) 2 CD DE 45 CI 45 DI GH CH F HI , and the highest compressive force occurs in 3 2 F (C) 2 E (1) Joint E EI DI (2) Joint D EI F (3) Joint I HI (4) Joint H members DE and AB = - . Thus (a) the Howe Pratt Bridge DC DC DH 45 45 DE BC EI EI CH DI E HI F (1) Joint E (2) Joint I (3) Joint D (4) Joint C BH 45 45 HI GH F (5) Joint H DI 45 bridge has the highest tensile force in a member, and (b) the value of the compressive force is the same in members DE and AB = 3 2 2 F (C) for both bridges. DE Problem 6.9 The truss shown is part of an airplane's internal structure. Determine the axial forces in members BC, BD, and BE. A 8 kN C E G 14 kN H 300 mm Solution: First, solve for the support reactions and then use the method of joints to solve for the reactions in the members. 8 kN 0.4 m 0.4 m 0.4 m 0.3 m BX BY 0.8 m FY B A B 400 mm 400 mm D 400 mm F 400 mm 14 kN 0.4 m 8 kN C E G 14 kN H 300 mm F 400 mm 400 mm Fx : Fy : + MB : Bx = 0 By + Fy - 8 - 14 = 0 (kN) (0.4)(8) + 0.8Fy - 1.2(14) = 0 D 400 mm 400 mm Solving, we get Bx = 0, By = 5.00 kN Fy = 17.00 kN. The forces we are seeking are involved at joints B, C, D, and E. The method of joints allows us to solve for two unknowns at a joint. We need a joint with only two unknowns. Joints A and H qualify. Joint A is nearest to the members we want to know about, so let us choose it. Assume tension in all members. Joint A: y Fx : Fy : -BC = 0 -AC + CE = 0 BC = 0, Solving, we get CE = 10.67 kN (T ) Joint B: y 8 kN AB BC AC AB 4 x BE 5 3 BD x sin = 0.6 cos = 0.8 = 36.87 Fx = AC + AB cos = 0 Fy = -8 - AB sin = 0 Solving, we get AC = 10.67 kN (T ) AB = -13.33 kN (C) Joint C: (Again, assume all forces are in tension) y BY We know AB = -13.33 kN BC = 0 By = 5.00 kN. We know 3 of the 5 forces at B Hence, we can solve for the other two. Fx : Fy : BD + BE cos - AB cos = 0 BC + By + BE sin + AB sin = 0 Solving, we get BD = -14.67 kN (C) BE = 5.00 kN (T ) CE x BC AC From Joint C, we had BC = 0 Thus BC = 0, BD = -14.67 kN (C) BE = 5.00 kN (T ) [AC = 10.67 kN (T )] Problem 6.10 For the truss in Problem 6.9, determine the axial forces in members DF , EF , and F G. Solution: First, solve for the support reactions and then use the method of joints to solve for the reactions in the members. 8 kN 0.4 m 0.4 m 0.4 m 0.4 m 300 mm 0.3 m BX BY Fx : Fy : + MB : Bx = 0 By + Fy - 8 - 14 = 0 (kN) (0.4)(8) + 0.8Fy - 1.2(14) = 0 8 kN A C E G 14 kN H 14 kN B 400 mm FY Joint G: D 400 mm 400 mm F 400 mm 0.8 m y EG G FG GH x Solving, we get Bx = 0, By = 5.00 kN Fy = 17.00 kN The forces we are seeking are involved with joints D, E, F , and G The method of joints allows us to solve for two unknown forces at a joint. We need to start with a joint with only two unknowns. Joints A and H qualify. Joint H is nearest to the members we want to know about, so let us start there. Assume all unknown forces are tensions. If we get a negative force in a solution, this will then imply compression in that member. Fx : Fy : GH - EG = 0 FG = 0 [GH = 18.67 kN (T )] Solving EG = 18.67 kN (T ) FG = 0 Joint F: y 14 kN y FG GH EF H FH x FH DF FY We know x Joint H 3 sin = = 0.6 5 cos = Fx : Fy : 4 = 0.8 5 -GH - F H cos = 0 -14 - F H sin = 0 Fx : Fy : F H = -23.33 kN (C) FG = 0 Fy = 17.00 kN F H cos - EF cos - DF = 0 F G + Fy + F H sin + EF sin = 0 Solving GH = 18.67 kN (T ), F H = -23.33 kN (C) Solving, we get EF = -5.00 kN (C) DF = -14.67 kN (C) and from above F G = 0 Problem 6.11 The loads F1 = F2 = 8 kN. Determine the axial forces in members BD, BE, and BG. D F1 3m B E 3m F2 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 B E A AX AY C 8m GY G 3m x D 3m F1 = 8 kN A C 4m 4m G F2 = 8 kN D F1 3m B E 3m F2 Fx : Fy : + MA : Ax + F1 + F2 = 0 (kN) Ay + Gy = 0 8Gy - 3F2 - 6F1 = 0 A C 4m Solving, G Solving for the external loads, we get Ax = -16 kN (to the left) Ay = -9 kN (downward) Gy = 9 kN (upward) Now use the method of joints to determine BD, BE, and BG. Start with joint D. Joint D: Joint E: 4m BD = 10 kN (T ) DE = -6 kN (C) y DE y BE D F1 = 8 kN DE x EG BD cos = 0.8 sin = 0.6 = 36.87 Fx : Fy : F1 - BD cos = 0 -BD sin - DE = 0 DE = -6 kN Fx = DE - EG = 0 Fy = -BE + F2 = 0 Solving: EG = -6 kN (C) BE = 8 kN (T ) F2 = 8 kN x 6.11 Contd. Joint G: Fx : -CG - BG cos = 0 BG sin + EG + Gy = 0 y Fy : Solving, we get BG EG BG = -5 kN (C) CG = 4 kN (T ) x CG GY Thus, we have BD = 10 kN (T ) BE = 8 kN (T ) BG = -5 kN (C) (EG = -6 kN (C)) Gy = 9 kN Problem 6.12 If the loads on the truss shown in Problem 6.11 are F1 = 6 kN and F2 = 10 kN, what are the axial forces in members AB, BC, and BD? Solution: Find the external support loads and then use the method of joints to determine loads in members. (Assume all loads in members to be tensions). External Loads: D y B 3m F2 = 10 kN F1 = 6 kN F1 D 3m B E 3m A C G F2 A AX AY 8m 3m x GY 4m 4m Joint A: y sin = 0.6 cos = 0.8 = 36.87 Fx : Fy : + MA : Ax + F1 + F2 = 0 Ay + Gy = 0 8Gy - 3F2 - 6F1 = 0 AX AB A AC x Solving, the external loads are Ax = -16 kN, Ay = -8.25 kN, Gy = 8.25 kN. Now use the method of joints to determine AB, BC, and BD. Start with Joint A: AY Ay = -8.25 kN Ax = -16 kN Fx : Fy : AC + Ax + AB cos = 0 Ay + AB sin = 0 6.12 Contd. Solving, AC = 5 kN (T ) AB = 13.75 kN (T ) Joint C: y BC AC C CG x (AC = 5 kN) Fx : Fy : Solving, BC = 0, CG = 5 kN (T ) Joint D: y F1 CG - AC = 0 BC = 0 BD DE x F1 = 6 kN Fx : Fy : F1 - BD cos = 0 -BD sin - DE = 0 Solving, we get DE = -4.5 kN (C) and BD = 7.5 kN (T ) Thus, we have AB = 13.75 kN (T ) BC = 0 BD = 7.5 kN (T ) Problem 6.13 The truss supports loads at C and E. If F = 3 kN, what are the axial forces in members BC and BE? 1m A B 1m D 1m Solution: The moment about A is 1m G C F 2F E MA = -1F - 4F + 3G = 0, from which G = 5 F 3 = 5 kN. The sums of forces: FY = AY - 3F + G = 0, from which AY = 4 F 3 = 4 kN. FX = AX = 0, from which AX = 0. The interior angles GDE, EBC are 45 , 1 from which sin = cos = . 2 Denote the axial force in a member joining I, K by IK. (1) Joint G: DG Fy = + G = 0, 2 from which 5 2 F = -5 2 kN (C). DG = - 2G = - 3 DG Fx = - - EG = 0, 2 from which 5 DG EG = - = F = 5kN (T ). 3 2 (2) Joint D: DG Fy = -DE - = 0, 2 from which DE = 5 F = 5 kN (T ). 3 1m A 1m B 1m D 1m G C F E 2F AY AX 1m F 1m DG 45 EG Joint G AY AB AC 45 Joint A G BD DG DE 45 Joint D 45 BC AC CE F Joint C 1m 2F 1m G DE BE 45 CE EG Joint E DG Fx = -BD + = 0, 2 from which 5 BD = - F = -5 kN (C). 3 (3) Joint E: BE Fy = - 2F + DE = 0, 2 from which BE = 2 2F - 2DE = 2 F 3 (4) Joint A: AC Fy = Ay - = 0, 2 = 2 kN (T ). from which AC = 4 2 F 3 = 4 2 kN (T ). BE Fx = -CE - + EG = 0, 2 from which 4 BE CE = EG - = F = 4 kN (T ). 3 2 AC Fx = AB + = 0, 2 from which AB = - 4 F = -4 kN (C). 3 (5) Joint C: AC Fy = BC + - F = 0, 2 from which BC = F - AC 2 = - 1 F = -1 kN (C). 3 Problem 6.14 Consider the truss in Problem 6.13. Each member will safely support a tensile force of 28 kN and a compressive force of 12 kN. Taking this criterion into account, what is the largest safe (positive) value of F ? Solution: From the solution to Problem 6.14, the member with the largest tensile force is EG = 5 F , from which F = 3 EG = 3 5 16.8 kN. The member with the largest compressive force is DG, DG = -5 2 F, 3 from which F = 5 3 DG 2 = 36 5 2 = 5.09 kN is the largest safe value. 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 360 lb at A and at B. What are the resulting axial forces in members CF , DF , and F G? Solution: Assume loads of 360 lbs at A and at B. Use the method of joints, starting with A and B, to work through the structure. Joint A: 1 ft G 1 ft F 1 ft 2 ft E D C 8 in y AC AX A x 360 lb 1 ft G B A 1 ft F 1 ft Fy : AC - 360 lb = 0 AC = 360 lb 2 ft Fx : Ax = 0 E D C 8 in If Ax = 0, then Bx = 0 because the stretcher must be in equilibrium Joint B: B BD BC A BX B 360 lb Fx : Fy : Bx + BC cos = 0 BC sin + BD - 360 = 0 Solving, BD = 360 lb, BC = 0 tan = 8 24 = 18.43 Bx = 0 6.15 Contd. Joint C: y CF CD C x AC BC tan = 2 1 = 63.43 BC = 0, AC = 360 lb Fx : Fy : -CD - BC cos - CF cos = 0 CF sin - BC sin - AC = 0 Solving, CD = -180 lb (C) CF = 402 lb (T ) Joint F: y F FG x DF CF = 63.43 (CF = 402 lbs (T )) Fx : Fy : CF cos - DF cos - F G = 0 -CF sin - DF sin = 0 Solving; we get DF FG and CF = -402 lb (C) = 360 lb (T ) from earlier = 402 lb (T ) Problem 6.16 Upon learning of an upgrade in the helicopter's engine, the engineer designing the truss shown in Problem 6.15 does new simulations and concludes that the downward forces the stretcher will exert at A and at B may be as large as 400 lb. What are the resulting axial forces in members DE, DF , and DG? Solution: Assume loads of 400 lb at A and B. Use the method of Joints, starting with A and B, and work through the structure. Joint A: y AY G 1 ft 1 ft F 1 ft 2 ft AX A F1 = 400 lb x E D C 8 in B A Fx : Fy : Ax = 0 Ay - F1 = 0 Fx : Fy : Bx + BC cos = 0 BC sin + BD - F2 = 0 F1 = 400 lb Solving, Ay = 400 lb. Ax = 0 If Ax = 0, then Bx = 0 for the stretcher not to move horizontally. (Ax + Bx = 0) Joint B: Solving, BC = 0, BD = 400 lb(T ) Joint C: CF y y CD BD BC C x AC BX B F2 x BC 2 1 tan = = 63.43 BC = 0, AC = 400 lb. Bx = 0 F2 = 400 lb 8 tan = 24 = 18.43 Fx : Fy : -CD - BC cos - CF cos = 0 CF sin - BC sin - AC = 0 Solving, CD = -200 lb (C) 6.16 Contd. Joint F: y FG F x CF DF = 63.43 CF = 44.7 lb (T ) Fx : Fy : CF cos - DF cos - F G = 0 -CF sin - DF sin = 0 Solving, we get CF = 447 lb (T ) DF = -447 lb (C) F G = 400 Joint D: y DG DF DE BD CD = -200 lb (C) BD = 400 lb (T ) DF = -447 lb (C) Fx : Fy : CD x CD - DE + DF cos - DG cos = 0 DF sin + DG sin - BD = 0 Solving DE = -968 lb (C) DG = 894 lb (T ) Thus, DE = -800 lb (C) DF = -447 lb (C) DG = 894 lb (T ) Problem 6.17 Determine the axial forces in the members in terms of the weight W . 1m A B E D 1m C W 0.8 m 0.8 m 0.8 m 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 = tan-1 0.8 = 38.66 . The angle formed by member DB with the positive x axis is 90 + . The angle formed by member AB with the positive x-axis is . Joint E: Fy = -DE cos - W = 0, from which DE = -1.28W (C) . Fy = -BE - DE sin = 0, from which BE = 0.8W (T ) Joint D: Fx = DE cos + BD cos - CD cos = 0, from which BD - CD = -DE. Fy = -BD sin + DE sin - CD sin = 0, from which BD + CD = DE. Solving these two equations in two unknowns: CD = DE = -1.28W (C) , BD = 0 Joint B: Fx = BE - AB sin - BD sin = 0, from which AB = BE sin = 1.28W (T ) Fy = -AB cos - BC = 0, from which BC = -AB cos = -W (C) Problem 6.18 Consider the truss in Problem 6.17. Each member will safely support a tensile force of 6 kN and a compressive force of 2 kN. Use this criterion to determine the largest weight W the truss will safely support. From the solution to Problem 6.17, the largest tensile force is in member AB, AB = 1.28W (T ), from which 6 W = 1.28 = 4.69 kN is the maximum safe load for tension. The largest compressive forces occur in members DE and CD, DE = CD = 1.28W (C), from which W = is the largest safe load for compression. 2 1.28 Solution: = 1.56 kN Problem 6.19 The loads F1 = 600 lb and F2 = 300 lb. Determine the axial forces in members AE, BD, and CD. F1 G D F2 6 ft C 3 ft E A 4 ft 4 ft B Solution: The reaction at E is determined by the sum of the moments about G: MG = +6E - 4F1 - 8F2 = 0, from which 4F1 + 8F2 = 800 lb. E = 6 The interior angle EAG is = tan -1 F1 G D F2 6 ft B 3 ft 4 ft A C 6 8 E = 36.87 . 4 ft 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 = E + AE = 0, from which AE = -E = -800 lb (C) . Fy = EG = 0, from which EG = 0. Joint A: Fy = -AE - AC cos = 0, from which AC = - AE = 1000 lb(T ). 0.8 F1 GX 6 ft GY F2 E EG E AE Joint E from which BD = 4 ft AB 4 ft BD BC F2 AB DG CD F1 AC AE Joint A F2 +AB 0.6 BD Joint D Joint B = -300 0.6 = -500 lb(C) . Fy = AC sin + AB = 0, from which AB = -AC(0.6) = -600 lb(C). Joint B: Fy = BD sin - AB - F1 = 0, Fx = -BC - BD cos = 0, from which BC = -BD(0.8) = 400 lb(T ). Joint D: Fy = -BD sin - CD - F1 = 0, from which CD = -F1 - BD(0.6) = -300 lb(C) Problem 6.20 Consider the truss in Problem 6.19. The loads F1 = 450 lb and F2 = 150 lb. Determine the axial forces in members AB, AC, and BC. Solution: From the solution to Problem 6.19 the angle = 36.87 and the reaction at E is E = 4F1 +8F2 = 500 lb. Denote the 6 axial force in a member joining two points I, K by IK. Joint E: Fy = EG = 0. Fx = AE + E = 0, from which AE = -E = -500 lb(C). Joint A: Fx = -AE - AC cos = 0, EG E Joint E AE AC AB Joint A AE BD BC F2 AB Joint B from which AC = - AE = 625 lb(T ) . 0.8 Fy = AC sin + AB = 0, from which AB = -AC(0.6) = -375 lb(C) Joint B: Fy = BD sin - F2 - AB = 0, from which BD = F2 +AB 0.6 = -375 lb(C) Fx = -BC - BD cos = 0, from which BC = -BD(0.8) = 300 lb(T ) Problem 6.21 Each member of the truss will safely support a tensile force of 4 kN and a compressive force of 1 kN. Determine the largest mass m that can safely be suspended. 1m 1m E 1m F 1m C 1m AB D m Solution: The common interior angle BAC = DCE = EF D = CDB is = tan-1 (1) = 45 . 1 Note cos = sin = . Denote the axial force in a member 2 joining two points I, K by IK. Joint F: DF Fy = - - W = 0, 2 from which DF = - 2W (C). Fx DF = -EF - = 0, 2 1m 1m E 1m F 1m C D 1m A B m from which EF = W (T ). Joint E: CE Fx = - + EF = 0 2 from which CE = 2W (T ). CE Fy = -ED - = 0, 2 from which ED = -W (C). Joint D: DF BD FY = ED + - = 0, 2 2 from which BD = -2 2W (C). DF BD FX = - - CD = 0, 2 2 from which CD = W (T ) Joint C: AC CE Fx = - + + CD = 0, 2 2 from which AC = 2 2W (T ) AC CE Fy = - + - BC = 0, 2 2 from which BC = -W (C) EF ED CE EF ED CD DF BD Joint D DF W Joint F CE CD BC Joint C Joint E BC BD AB B Joint B AC Joint B: BD Fx = -AB + = 0, 2 from which AB = -2W (C) This completes the determination of the axial forces in nine members. The all maximum tensile force occurs in member AC, AC = 2 2W (T ), from which 4 the safe load is W = = 2 = 1.414 kN. The maximum compression 2 2 occurs in member BD, BD = -2 2W (C), from which the maximum safe 1 load is W = = 0.3536 kN. The largest mass m that can be safely 2 supported is m = 2 353.6 9.81 = 36.0 kg 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? A B D F H 2m C 6m 6m E 6m G 6m I 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. 50 kN 6m AY 3m 50 kN 6m 50 kN 6m 3m x IY 50 kN B D F H 2m A 6m C 6m E 6m G 6m I Fy : MA : Ay + Iy - 4(50) = 0 (kN) -3(50) - 9(50) - 15(50) - 21(50) + 24 Iy = 0 AB = -180.3 kN = 33.69 Fx : Fy : BC cos + BD - AB cos = 0 -50 - AB sin - BC sin = 0 Solving Ay = 100 kN Iy = 100 kN Joint A: y AB Solving, BC = 90.1 kN (T ) BD = -225 kN (C) Joint C: A AC AY y x BC CD 2 3 C CE x tan = AC = 33.69 = 33.69 Fx : Fy : AB cos + AC = 0 AB sin + Ay = 0 AC = 150 kN (T ) BC = 90.1 kN (T ) Fx : CE - AC + CD cos - BC cos = 0 CD sin + BC sin = 0 Solving, AB = -180.3 kN (C) AC = 150 kN (T ) Joint B: y 50 kN B Fy : Solving, CE = 300 kN (T ) CD = -90.1 kN (C) Hence BD AB BC x BC = 90.1 kN (T ) CD = -90.1 kN (C) CE = 300 kN (T ) Problem 6.23 For the Warren truss in Problem 6.22, determine the axial forces in members DF , EF , and F G. 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: y 50 kN BD D A DF 6m x Solving, we get EF = 0 EG = 300 kN (T ) Note: The results are symmetric to this point! Joint F: B D F H 2m C 6m E 6m G 6m I CD = 33.69 BD = -225 kN (C) CD = -90.1 kN (C) Fx : Fy : Solving, DE y 50 kN DF - BD + DE cos - CD cos = 0 -50 - CD sin - DE sin = 0 DF = -300 kN (C) DE = 0 DF F FH x 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: EF = 33.69 DF = -300 kN (C) EF = 0 FG y DE EF Fx : Fy : F H - DF + F G cos - EF cos = 0 -50 - EF sin - F G sin = 0 CE = 33.69 CE = 300 kN (T ) DE = 0 Fx : Fy : E EG x Solving: F H = -225 kN (C) F G = -90.1 kN (C) Thus, we have DF = -300 kN (C) EF = 0 F G = -90.1 kN (C) Note-symmetry holds! EG - CE + EF cos - DE cos = 0 DE sin + EF sin = 0 Problem 6.24 The Pratt bridge truss supports five forces (F = 300 kN). The dimension L = 8 m. Determine the axial forces in members BC, BI, and BJ. L A L B L C L D L E L G L H I F F J F K L F M F Solution: diagram, Find support reactions at A and H. From the free body Fx = AX = 0, Fy = AY + HY - 5(300) = 0, and MA = 6(8)HY - 300(8 + 16 + 24 + 32 + 40) = 0. L B L A I L C L D L E L G L H J K L M From these equations, AY = HY = 750 kN. From the geometry, the angle = 45 Joint A: From the free body diagram, Fx = AX + TAB cos + TAI = 0, Fy = TAB sin + AY = 0. From these equations, TAB = -1061 kN and TAI = 750 kN. Joint I: From the free body diagram, Fx = TIJ - TAI = 0, Fy = TBI - 300 = 0. From these equations, TBI = 300 kN and TIJ = 750 kN. Joint B: From the free body diagram, Fx = TBC + TBJ cos - TAB cos = 0, Fy = -TBI - TBJ sin - TAB sin = 0. From these equations, TBC = -1200 kN and TBJ = 636 kN. B A AY G J L 8 L 8 F K L 8 L L 8 M L 8 H L I L 8 HY F F L=8m Joint A y TAB A TAI AY F F F = 300 kN Joint I y TBI I x TAI TIJ F x Joint B y TBC TAB TBI x TBJ Problem 6.25 For the Pratt bridge truss in Problem 6.24, determine the axial forces in members CD, CJ, and CK. Solution: Use all of the known values from Problem 6.24, and start with Joint J. Joint J: From the free body diagram, Fx = TJK - TBJ cos - TIJ = 0, Fy = TCJ + TBJ sin - 300 = 0. From these equations, TJK = 1200 kN and TCJ = -150 kN. Joint C: From the free body diagram, Fx = TCD + TCK cos - TBC = 0, Fy = -TCJ - TCK sin = 0. From these equations, TCD = -1350 kN and TCK = 212 kN. Joint J y TBJ Joint C y TBC x TCJ TCD x TIJ TCJ TJK F TCK Problem 6.26 The Howe truss helps support a roof. Model the supports at A and G as roller supports. Determine the axial forces in members AB, BC, and CD. 400 lb 800 lb 600 lb D 400 lb C B A H 4 ft 4 ft I 4 ft J 4 ft K 4 ft L 4 ft E 8 ft F G 600 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 = tan -1 800 lb 600 lb 400 lb 8 ft B A H I J K L C D 600 lb E 400 lb F G 8 12 = 33.7 . The length of the vertical members: BH = 4 8 12 = 2.6667 ft, from which the angle HIB = tan-1 CI = 8 2.6667 4 = 33.7 . 4 ft 4 ft 4 ft 4 ft 4 ft 4 ft 8 = 5.3333 ft, 12 5.333 4 800 lb 600 lb 400 lb A 4 ft 4 ft 4 ft 4 ft 4 ft 4 ft AB 1400 lb Joint A BI Pitch CI Pitch from which the angle IJC = tan-1 = 53.1 . 600 lb 400 lb G The moment about G: MG = (4 + 20)(400) + (8 + 16)(600) + (12)(800) - 24A = 0, from which A = 33600 = 1400 lb. Check: The total load is 2800 lb. 24 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 = AB sin P + 1400 = 0, 1400 from which AB = - sin = -2523.9 lb (C) p BH AH HI 400 lb Pitch AB AH BC Pitch Pitch Joint H 600 lb CD BH BI Joint B HI IJ Joint I Joint H: IJC BC CI CJ Joint C Fx = AB cos Pitch + AH = 0, from which AH = (2523.9)(0.8321) = 2100 lb (T ) Fy = BH = 0, or, BH = 0. Fx = -AH + HI = 0, from which HI = 2100 lb (T ) Joint B: Fx = -AB cos Pitch + BC cos Pitch +BI cos Pitch = 0, from which BC + BI = AB 6.26 Contd. Fy = -400 - AB sin Pitch + BC sin Pitch -BI sin Pitch = 0, from which BC - BI = AB + 400 . sin Pitch Solve the two simultaneous equations in unknowns BC, BI: 400 = -360.56 lb (C), 2 sin Pitch BI = - and BC = AB - BI = -2163.3 lb (C) Joint I: Fx = -BI cos Pitch - HI + IJ = 0, from which IJ = 1800 lb (T ) Fy = +BI sin Pitch + CI = 0, from which CI = 200 lb (T ) Joint C: Fx = -BC cos Pitch + CD cos Pitch + CJ cos IJC = 0, from which CD(0.8321) + CJ(0.6) = -1800 Fy = -600 - CI - BC sin Pitch + CD sin Pitch -CJ sin IJC = 0, from which CD(0.5547) - CJ(0.8) = -400 Solve the two simultaneous equations to obtain CJ -666.67 lb (C), = and CD = -1682.57 lb (C) 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 = 45 . Determine the axial forces in the members of the truss. B 1.8 m 2.2 m F A C D H E G 3.4 m 3.4 m 3.4 m 3.4 m Solution: = tan-1 = tan-1 = tan-1 The included angles 4 3.4 2.2 3.4 1.8 3.4 = 49.64 , = 32.91 , B 1.8 m C F A G H 2.2 m D E = 27.9 . The complete structure as a free body: The sum of the moments about A is MA = -(75)(3.4)(1 + 2 + 3) + (4)(3.4)Ey = 0. with this relation and the fact that Ex cos 45 + Ey cos 45 = 0, we obtain Ex = -112.5 kN and Ey = 112.5 kN. From A Fx = Ax + Ex = 0, AX = -EX = 112.5 kN. A Fy = Ay - 3(75) + Ey = 0, A 3.4 m 3.4 m 3.4 m 3.4 m 75 kN 75 kN 75 kN AX AY 3.4 m 3.4 m BF EX 3.4 m 3.4 m EX EY from which Ay = 112.5 kN. Thus the reactions at A and E are symmetrical 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 = AB cos + Ax + AF cos = 0, Fy = AB sin + Ay + AF sin = 0, from which two simultaneous equations are obtained. Solve: and Joint E: Fy = -DE cos + Ex - EH cos = 0. Fy = DE sin + Ey + EH sin = 0, from which two simultaneous equations are obtained. AF = -44.67 kN (C) , AB = -115.8 kN (C) AB AX AY Joint A 75 kN BC BF BG AB Joint B EH AF DE DH FG GH CD DG EY Joint E 75 kN AF EH DH Joint D Joint F 75 kN BC CD CG Joint C Joint H DE Solve: and Joint F: EH = -44.67 kN(C) , DE = -115.8 kN(C) Fx = -AF cos + F G = 0, from which F G = -37.5 kN (C) Fy = -AF sin + BF = 0, from which BF = -24.26 kN (C) Joint H: Fx = EH cos - GH = 0, 6.27 Contd. from which GH = -37.5 kN (C) Fy = -EH sin + DH = 0, from which DH = -24.26 kN (C) Joint B: Fy = -AB sin - BF + BG sin - 75 = 0, from which BG = 80.1 kN (T ) Fx = -AB cos + BC + BG cos = 0, from which BC = -145.8 kN (C) Joint D: Fy = -DE sin - DH - DG sin - 75 = 0, from which DG = 80.1 kN (T ) Fx = DE cos - CD - DG cos = 0, from which CD = -145.8 kN (C) Joint C: Fx = CD - BC = 0, from which CD = BC Check. Fy = -CG - 75 = 0, from which CG = -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) 1000 lb C 4 ft A B D 2 ft 2000 lb 5 ft 5 ft 5 ft Solution: Problem 6.28 don't have unique solution Problem 6.29 (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). 2m 3m A 2m B C 2 kN D 1m 2 kN Solution: Problem 6.29 don't have unique solution Problem 6.30 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 1m E B C D F G H J 100 kN Solution: DJ 45 (a) Start with Joint J A 1m J HJ 100 kN Fx : Fx : Fy : Solving -HJ - DJ cos 45 = 0 DJ sin 45 - 100 = 0 DJ = 141.4 kN (T ) HJ = -100 kN (C) Joint H: B C D E 1m F 1m G 1m H 1m J 100 kN -CD - DG cos 45 + DJ cos 45 = 0 -DG sin 45 - DJ sin 45 - DH = 0 Fy : Solving, CD = 200 kN DG = -141.4 kN (C) (b) Method of Sections CD D 45 DH DG J HJ GH H 1m 100 kN 1m GH Fx : Fy : H HJ - GH = 0 DH = 0 DH = 0, GH = -100 kN (C) Fx : Fy : MD : -CD - DG cos 45 - GH = 0 -DG sin 45 - 100 = 0 -(1)GH - (1)(100) = 0 Joint D CD D x 45 Solving, GH = -100 kN (C) CD = 200 kN (T ) DG = -141.4 kN (C) 45 DJ DH DG Problem 6.31 For the truss in Problem 6.30, use the method of sections to determine the axial forces in members BC, CF , and F G. Solution: BC 45 CF J F FG G 1m H 1m 100 kN 1m C A 1m E B C D F G H J D 100 kN Solving BC = 300 kN (T ) CF = -141.4 kN (C) F G = -200 kN (C) Fx : Fy : MC : -BC - CF cos 45 - F G = 0 -CF sin 45 - 100 = 0 -(1)F G - 2(100) = 0 Problem 6.32 Use the method of sections to determine the axial forces in members AB, BC, and CE. A 1m B 1m D 1m Solution: AX AY First, determine the forces at the supports B D = 45 1m G C 1m F 1m E 1m 2F GY 1m C F E 2F Fx : Fy : + MA : Ax = 0 Ay + Gy - 3F = 0 -1(F ) - 2(2F ) + 3Gy = 0 1m A B 1m D 1m 1m Solving Ax = 0 Gy = 1.67F Ay = 1.33F Method of Sections: G C F y AY = 1. 33 F AX = 0 2F Fx : Fy : CE + AB = 0 BC + Ay - F = 0 (-1)Ay + (1)CE = 0 E AX = 0 AB B BC AY 1m + MB : Solving, we get 1m C F CE x AB = -1.33F (C) CE = 1.33F (T ) BC = -0.33F (C) Problem 6.33 The truss supports loads at A and H. Use the method of sections to determine the axial forces in members CE, BE, and BD. A 18 kN C E G 24 kN H 300 mm B D 400 mm 400 mm F 400 mm 400 mm Solution: 18 kN First find the external support loads on the truss 18 kN 24 kN C E G H 0.8 m BX BY 0.4 m 1.2 m Fx : Fy : MB : Solving: Bx = 0 By + Fy - 18 - 24 = 0 (kN) 0.8Fy - (1.2)(24) + (0.4)18 = 0 Solving, Bx = 0 By = 15 kN Fy = 27 kN Method of sections: + Fx : Fy : MB : CE + BE cos + BD = 0 By - 18 + BE sin = 0 +(0.4)(18) - (0.3)(CE) = 0 CE = 24 kN (T ) BE = 5 kN (T ) BD = -28 kN (C) 24 kN C E G H 300 mm A B D FY 400 mm 400 mm 400 mm F 400 mm D 18 kN C CE BE E 0.4 m BD 0.3 m BY tan = 3 4 = 36.87 By = 15 kN Problem 6.34 For the truss in Problem 6.33, use the method of sections to determine the axial forces in members EG, EF , and DF . Solution: at B and F are Bx = 0, By = 15 kN, Fy = 27 kN. From the solution to Problem 6.33, the external forces 18 kN A C E G 24 kN H 300 mm B D 400 mm 400 mm 400 mm F 400 mm 24 kN E 0.3 m D EG EF G H DF F FY 0.4 m = 36.87 Fx : Fy : MF : -EG - EF cos - DF = 0 -24 + Fy + EF sin = 0 -(0.4)(24) + (0.3)EG EG = 32 kN (T ) EF = -5 kN (C) DF = -28 kN (C) Solving: Problem 6.35 For the Howe and Pratt trusses, use the method of sections to determine the axial force in member BC. A L B L C L D L L E G F H 2F Howe L B L C L D L A G F H 2F Pratt I 2F E L I 2F Solution: From the free body diagram of the whole truss, the equations of equilibrium are Fx = AX = 0, Fy = AY + EY - 5F = 0, and MA = 4LEY - LF - (2L)2F - (3L)2F = 0. L B L C L D L A G F H 2F Howe L B A G F H 2F Pratt I 2F C L D I 2F E L From these equations, we get AX = 0, AY = 2.25F, and EY = 2.75F . Note that the support forces are the same for the Howe and Pratt trusses. Howe Section: From the Howe truss section, we see that if we sum moments about G, we get one equation in one unknown, i.e., MG = -LAY - LTBCHowe = 0, or TBCHowe = -2.25F (compression). Pratt Section: From the Pratt truss section we see that summing moments about H is advantageous. Hence, MH = -2LAY + LF - LTBCPratt = 0, or TBCPratt = -3.5F (compression). L L L E y L AX A AY B L G F C L H 2F D L I 2F TBC TBH AX AY G T H x GH PRATT L E EY x y AX AY B G F HOWE TBC TGC TGH x y B Problem 6.36 For the Howe and Pratt trusses in Problem 6.35, determine the axial force in member HI. Solution: Howe Section: From the Howe truss section, we see that if we sum moments about C, we get one equation in one unknown, i.e., MC = 2LEY - 2LF - LTHIHowe = 0, or THIHowe = 3.5F (tension). Pratt Section: From the Pratt truss section we see that summing moments about D is advantageous. Hence, MD = LEY - LTHIPratt = 0 or THIPratt = 2.75F (tension). TCD C TCD D TCI I H THI 2F HOWE y E x EY D y E EY x THD I H THI 2F PRATT Problem 6.37 The Pratt bridge truss supports five forces F = 340 kN. The dimension L = 8 m. Use the method of sections to determine the axial force in member JK. L A L B L C L D L E L G L H I F F J F K L F M F Solution: First determine the external support forces. L L B L L F F L F L F L F L L C D L E L G L AX AY L HY A I J F K F F L M F H F = 340 kN, L = 8 M F Fx : Fy : + Solving: MA : Ax = 0 Ay - 5F + Hy = 0 6LHy - LF - 2LF - 3LF - 4LF - 5LF = 0 = 45 L = 8M F = 340 kN Ay = 850 kN Fx : Fy : + MC : CD + JK + CK cos = 0 Ay - 2F - CK sin = 0 L(JK) + L(F ) - 2L(Ay ) = 0 Ax = 0, Ay = 850 kN Hy = 850 kN Note the symmetry: Method of sections to find axial force in member JK. B C CD D CK J L K Solving, JK = 1360 kN (T ) Also, CK = 240.4 kN (T ) CD = -1530 kN (C) A L AY F F I JK Problem 6.38 For the Pratt bridge truss in Problem 6.41, use the method of sections to determine the axial force in member EK. Solution: From the solution to Problem 6.37, the support forces are Ax = 0, Ay = Hy = 850 kN. Method of Sections to find axial force in EK. DE E G L A I L F F HY F F J F K F L M F H L B L C L D L E L G L EK KL Solution: EK = 240.4 kN (T ) Also, KL = 1360 kN (T ) DE = -1530 kN (C) Fx : Fy : ME : -DE - EK cos - KL = 0 Hy - 2F - EK sin = 0 -(L)(KL) - (L)(F ) + (2L)Hy = 0 Problem 6.39 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. B D F H 2m A 6m C 6m E 6m G 6m I Solution: First, find the external support forces. By symmetry, Ay = Iy = 100 kN (we solved this problem earlier by the method of joints). y A B 50 kN BD 2m 6m AY Solving: CE = 300 kN (T ) 2 tan = 3 = 33.69 Fx : Fy : MC : BD + CD cos + CE = 0 Ay - 50 + CD sin = 0 -6Ay + 3(50) - 2BD = 0 Also, BD = -225 kN (C) CD = -90.1 kN (C) D B CD CE x D F H 2m A 6m C 6m E 6m G 6m I C Problem 6.40 The walkway in Problem 6.39 exerts equal vertical loads on the Warren truss at B, D, F , and H. Use the method of sections to determine the maximum allowable value of each vertical load if the magnitude of the axial force in member F G is not to exceed 100 kN. B D F H 2m A 6m C 6m E 6m G 6m I Solution: Let the loads at B, D, F , and H be denoted by W . By summetry Ay = Iy = 2W . Method of Sections W F FG FH H EG G 3m 2m 3m I IY tan = 2 3 = 33.69 Fx : Fy : MG : -EG - F H - F G cos = 0 Iy - W + F G sin = 0 2F H + 6Iy - 3W = 0 We set F G = 100 kN and solve: For F G = +100 kN, W = -55.5 kN (this implies an upward load on the bridge) For F G = -100 kN (in compression) W = 55.5 kN. This is the load limit on the bridge based on the load in member F G. Problem 6.41 The mass m = 120 kg. Use the method of sections to determine the axial forces in members BD, CD, and CE. 1m 1m 1m E 1m F C 1m AB D m Solution: First, find the support reactions using the first free body diagram. Then use the section shown in the second free body diagram to determine the forces in the three members. Support Reactions: Equilibrium equations are Fx = AX = 0, Fy = AY + BY - mg = 0, and summing moments around A, MA = -3mg + (1)BY = 0. Thus, AX = 0, AY = -2.35kN, and BY = 3.53 kN Section: From the second free body diagram, the equilibrium equations for the section are Fx = AX + TCD + TCE cos(45 ) + TBD cos(45 ) = 0, Fy = AY + BY + TCE sin(45 ) + TBD sin(45 ) = 0, and, summing moments around C, MC = (1)TBD cos(45 ) - (1)AY + (1)AX = 0. Solving, we get TBD = -3.30 kN, TCD = 1.18 kN, TCE = 1.66 kN. AY AX 1m 1m C 1m AX AY B 1m E 1m F 120 g D BY 1m 1m C 1m 1m TCE E 1m F TCD TBD B BY D Problem 6.42 For the truss in Problem 6.41, use the method of sections to determine the axial forces in members AC, BC, and BD. Solution: Use the support reactions found in Problem 6.41. The free body diagram for the section necessary to find the three unknowns is shown at right. The equations of equilibrium are Fx = AX + TAC cos(45 ) + TBD cos(45 ) = 0, Fy = AY + BY + TBC + TAC sin(45 ) + TBD sin(45 ) = 0, and, summing moments around B, MB = (-1)AY - (1)TAC sin(45 ) = 0. The results are TAC = 3.30 kN, TBC = -1.18 kN, and TBD = -3.30 kN. 1m 1m 1m AX AY TAC C 1m E 1m F TBC D TBD BY Problem 6.43 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 kN 2 kN D 2 kN C B A H 2 m I 2m J 2m K 2m L 2m 2m E 4m F G 2 kN 2 kN Solution: = tan-1 The pitch of the roof is 4 6 = 33.69 . 2 kN 2 kN 2 kN B A H I J K L C D 2 kN E 2 kN F 4m G 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 = -2(2)(1 + 2 + 3 + 4 + 5) + 6(2)G = 0, from which G = 30 6 = 5 kN. The sum of the forces: FY = A - 5(2) + G = 0, from which A = 10 - 5 = 5 kN. The method of joints: Denote the axial force in a member joining I, K by IK. (a) Joint A: Fy = A + AB sin = 0, from which AB = -A sin 2m 2m 2m 2m F F 2m 2m F F = 2 kN G = -5 0.5547 = -9.01 kN (C). F Fx = AB cos + AH = 0, from which AH = -AB cos = 7.5 kN (T ). Joint H: Fy = BH = 0. Joint B: Fx = -AB cos + BI cos + BC cos = 0, Fy = -2 - AB sin - BI sin + BC sin = 0. Solve: BI = -1.803 kN (C) , BC = -7.195 kN (C) Make the cut through BC, BI and HI. The section as a free body: The sum of the moments about B: A 2m 2m 2m 2m 2m 2m AB (a) A BH AH HI 2 kN AH AB BC BI Joint A (b) Joint H F B BH Joint B BC BI HI MB = -A(2) + HI(2 tan ) = 0, from which HI = 3 A 2 (b) A = 7.5 kN(T ). The sum of the forces: Fx = BC cos + BI cos + HI = 0, Fy = A - F + BC sin - BI sin = 0. Solve: BI = -1.803 kN (C) . 2m Problem 6.44 Consider the truss in Problem 6.43. Use the method of sections to determine the axial force in member EJ. Solution: From the solution to Problem 6.43, the pitch angle is = 36.69 , and the reaction G = 5 kN. The length of member EK is 16 = 2.6667 m. 6 DE EJ JK F E F LEK = 4 tan = 2m 2m G The interior angle KJE is LEK 2 = tan-1 = 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 = +4G - 2(F ) - JK(2.6667) = 0, from which JK = 20-4 2.6667 = 6 kN (T ). The sum of the forces: Fx = -DE cos - EJ cos - JK = 0. Fy = DE sin - EJ sin - 2F + G = 0, from which the two simultaneous equations: 0.8321DE + 0.6EJ = -6, 0.5547DE - 0.8EJ = -1. Solve: EJ = -2.5 kN (C) . Problem 6.45 Use the method of sections to determine the axial force in member EF . 10 kip A 4 ft 10 kip B C 4 ft D E 4 ft F G 4 ft H I 12 ft Solution: = tan-1 The included angle at the apex BAC is 12 16 = 36.87 . 10 kip A 4 ft The interior angles BCA, DEC, F GE, HIG are = 90 - = 53.13 . The length of the member ED is LED = 8 tan = 6 ft. The interior angle DEF is = tan -1 10 kip B D C 4 ft E 4 ft 4 LED = 33.69 . The complete structure as a free body: The moment about H is MH = -10(12) - 10(16) + I(12) = 0, from which I = 280 = 23.33 kip. 12 The sum of forces: Fy = Hy + I = 0, from which Hy = -I = -23.33 kip. Fx = Hx + 20 = 0, from which Hx = -20 kip. Make the cut through EG, EF , and DE. Consider the upper section only. 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 = -10(4) - 10(8) + DF (LED ) = 0, from which DF = 120 = 20 kip. 6 The sum of forces: Fy = -EF sin - EG sin - DF = 0, Fx = -EF cos + EG cos + 20 = 0, from which the two simultaneous equations: 0.5547EF + 0.8EG = -20, and 0.8320EF - 0.6EG = 20. Solve: EF = 4.0 kip (T ) F G 4 ft H I 12 ft F = 10 kip F = 10 kip E EG DF EF Problem 6.46 Consider the truss in Problem 6.45. Use the method of sections to determine the axial force in member F G. Solution: From the solution of Problem 6.45, the apex A included angle is = 36.87 . The length of the member F G is LF G = 12 tan = 9 ft. Make the cut through EG, GF , and F H, and consider the upper section. Denote the axial force in a member joining I, K by IK. The section as a free body: The cut in EG is made very near the point G; the moment about this cut by MG = -(8 + 12)F + LF G F H = 0 (where F = 10 kip from Problem 6.45), from which F H = 22.22 kip (T ). The sum of the forces, from which EG = -27.78 kip(C). Fx = F G + 2F + EG cos = 0, from which F G = -3.33 kip (C) . F = 10 F = 10 EG 9 ft FG FH G Problem 6.47 The load F = 20 kN and the dimension L = 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 freebody diagram is statically indeterminate. Solution: The complete structure as a free body: The sum of the moments about K is MK = -F L(2 + 3) + M L(2) = 0, from which M = 5F = 50 kN. The sum of forces: 2 FY = KY + M = 0, from which KY = -M = -50 kN. FX = KX + 2F = 0, from which KX = -2F = -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 = Kx - HI + IJ = 0, from which HI - IJ = Kx . Sum moments about K to get MK = M (L)(2) + JM (L)(2) - IJ(L) + HI(L) = 0. Substitute HI - IJ = Kx , to obtain JM = -M - Kx = 2 -30 kN (C). Fy = Ky + M + JM + HK = 0, from which HK = -JM = 30 kN(T ) L A F B L C L F D E G L I H J L K M L A B L C L G I L J L D E H K M F 2L L F 2L KX KY HI HJ M HK KX KY 2L JM M L Problem 6.48 The weight of the bucket is W = 1000 lb. The cable passes over pulleys at A and D. (a) Determine the axial forces in member F G and HI. (b) By drawing free-body diagrams of sections, explain why the axial forces in members F G and HI are equal. 3 ft 3 ft 3 in L 3 ft K J D C F H E G I 35 W B 3 ft 6 in A The truss is at angle = 35 relative to the horizontal. The angles of the members F G and HI relative to the horizontal are = 45 + 35 = 80 . (a) Make the cut through F H, F G, and EG, and consider the upper section. Denote the axial force in a member joining, , by . The section as a free body: The perpendicular distance from point F is LF W = 3 2 sin + 3.5 = 7.678 ft. The sum of the moments about F is MF = -W LF W + W (3.25) - |EG|(3) = 0, from which EG = -1476.1 lb (C). The sum of the forces: Solution: D C F H 3 ft 3 ft 3 in L I 35 K J G E B 3 ft 6 in A FY = -F G sin - F H sin - EG sin - W sin - W = 0, FX = -F G cos - F H cos - EG cos - W cos = 0, from which the two simultaneous equations: -0.9848F G - 0.5736F H = 726.9, and -0.1736F G - 0.8192F H = -389.97. Solve: F G = -1158.5 lb (C) , and F H = 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 = 3 cos + 3 2 sin + 3.5 = 10.135 ft. The sum of the moments about H is MH = -W (L) - |GI|(3) + W (3.25) = 0, from which |GI| = -2295 lb (C). FY = -HI sin - JH sin - GI sin - W sin - W = 0, FX = -HI cos - JH cos - GI cos - W cos = 0, from which the two simultaneous equations: -0.9848HI - 0.5736JH = 257.22, and -0.1736HI - 0.8192JH = -1060.8. Solve: and HI = -1158.5 lb(C) , JH = 1540.6 lb(T ) . 3 ft W FH 3.25 ft 3 ft FG EG W 3.5 ft W W JH HI GI (b) Choose a coordinate system with the y-axis parallel to JH. Isolate a section by making cuts through F H, F G, 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 F G and HI, the two axial forces must be equal: Fx = HI cos 45 - F G cos 45 = 0, from which HI = F G Problem 6.49 Consider the truss in Problem 6.48. The weight of the bucket is W = 1000 lb. The cable passes over pulleys at A and D. Determine the axial forces in members IK and JL. JL 3.25 ft Make a cut through JL, JK, and IK, and consider the upper section. Denote the axial force in a member joining, , by . The section as a free body: The perpendicular distance from point J to the line of action of the weight is L = 6 cos + 3 2 sin + 3.5 = 12.593 ft. The sum of the moments about J is MJ = -W (L) + W (3.25) - IK(3) = 0, from which IK = -3114.4 lb(C). The sum of the forces: Fx = JL cos - IK cos -W cos - JK cos = 0, and Fy = -JL sin - IK sin -W sin - W - JK sin = 0, and and W W 3.5 ft Solution: 3 ft JK IK from which two simultaneous equations: 0.8192JL + 0.1736JK = -1732 0.5736JL + 0.9848JK = 212.75. JL = 2360 lb(T ) , JK = -1158.5 lb(C) . Solve: Problem 6.50 The truss supports loads at N , P , and R. Determine the axial forces in members IL and KM . 1m 2m 2m 2m 2m 2m K J 2m M L O N 1 kN Q P 2 kN I H G F E D C B R 1 kN 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, , by . The section as a free body: The moment about M is MM = -IL - 2(1) - 4(2) - 6(1) = 0, from which IL = -16 kN (C) . tan-1 (0.5) = 26.57 . Solution: 2m 2m 2m A The angle of member IM is = The sums of the forces: 6m Fy = -IM sin - 4 = 0, 4 from which IM = - sin = -8.944 kN (C). Fx = -KM - IM cos - IL = 0, from which KM = 24 kN(T ) KM IM IL 1 kN 2 kN 2m 1 kN 2m 2m 1m 2m 2m 2m 2m 2m 1m 2m 2m 2m 2m A H F D E C B K J I G 1 kN 2 kN 1 kN L M O N Q P R 6m Problem 6.51 Consider the truss in Problem 6.50. 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 = tan-1 4 = 26.57 . The moment about B is 8 MB = 6AJ cos - 24 = 0, from which AJ = 4.47 kN (T ). The section as a free body: The angles of members HJ and HI relative to the vertical are = tan-1 2 = 14.0 , and = 8 tan-1 1.5 = 36.87 respectively. Make a cut the through four 2 members AJ, HJ, HI, and GI, and consider the upper section. The moment about the point I is MI = -24+2AJ cos +2HJ cos = 0. From which HJ = 8.25 kN (T ) . The sums of the forces: Fx = -AJ sin + HJ sin - HI sin = 0, from which HI = AJ sin -HJ sin sin 1m AJ HJ HI 2m I GI 2m 1 kN 2m 2 kN 2m 1 kN 2m = 2-2 sin = 0. FY = -AJ cos - HJ cos - HI cos - GI - 4 = 0, from which GI = -16 kN (C) Problem 6.52 Determine the reactions on member AB at A. (Notice that BC is a two-force member.) A 200 N B 400 mm C 300 mm 300 mm 400 mm Solution: Since BC is a two force member, the force in BC must be a long the line between B and C. 200 N AX 0.3 m AY 0.4 m 45 0.4 m Fx : Fy : MA : Solving: Ax + FBC cos 45 = 0 Ay - FBC sin 45 - 200 = 0 -(0.3)(200) - (0.6)CFBC sin 45 = 0 Ax = 100 N, Ay = 100 N FBC = 141.2 N (compression) 200 N A B y B FBC x 0.3 m 400 mm C 300 mm 300 mm 400 mm Problem 6.53 (a) Determine the forces and couples on member AB for cases (1) and (2). (b) You know that the moment of a couple is the same about any point. Explain why the answers are not the same in cases (1) and (2). 200 N-m A B C 1m (1) 1m 200 N-m A B C 1m (2) 1m Solution: Case (a) Element BC: The moment about B is MB = (1)Cy = 0, hence Cy = 0 . The sum of the forces: Fy = By + Cy = 0, from which By = 0 . 200 N-m A B C 1m Fx = Bx = 0. Element AB: The sum of the moments about A: M = MA -200 = 0, from which MA = 200 N-m . The sum of forces: Fy = By + Ay = 0, from which Ay = 0 , Fx = Ax = 0. Case (b) Element BC: The sum of the moments about B: MB = (1)Cy - 200 = 0, from which Cy = 200 N. The sum of the forces: BC Fy = By + Cy = 0, BC from which By = -200 N. BC Fx = Bx = 0. 1m (a) A B 200 N-m C 1m (b) AY (a) AX AB 1m BY BY BY BC (b) AX AB BC MA 200 N-M BX BX AB AY BY AB BC BX BX MA BC CY 200 N-M CY Element AB: The moments about A: AB M = MA + (1)By = 0, from which, since the reactions across the joint are equal and opposite: AB BC By = -By = 200 N , MA = -200 N-m . The sum of the forces: AB Fy = Ay + By = 0, from which Ay = -200 N . AB Fx = Ax + Bx = 0, from which Ax = 0. Explanation of difference: The forces are equal and opposite across the joint B, so it matters on which side of B the couple is applied. Problem 6.54 For the frame shown, determine the reactions at the built-in support A and the force exerted on member AB at B. 6 ft A B 200 lb C 20 6 ft 3 ft 3 ft Solution: Element AB: The equilibrium equations are: FX = AX + BX = 0, FY = AY + BY = 0, and MA = NA + (6)BY = 0. FX = -BX - C sin(20 ) = 0, FY = -BY - 200 + C cos(20 ) = 0, and, summing moments around B, MB = -(3)200 - (6)C sin(20 ) + (6)C cos(20 ) = 0. We have six equations in six unknowns. Solving simultaneously yields AX = 57.2 lb, AY = 42.8 lb, BX = -57.2 lb, BY = -42.8 lb, C = 167.3 lb, and NA = 256.6 ft-lb. A 6 ft B 200 lb Element BC: The equilibrium conditions are C 6 ft 3 ft 3 ft 20 NA AX 6 ft AY A BY B BX 6 ft BY BX 6 ft C 6 ft 3 ft 3 ft C 20 20 B 200 lb Problem 6.55 The force F = 10 kN. Determine the forces on member ABC, presenting your answers as shown in Fig. 6.35. F D E G A 1m B 1m C 2m 1m Solution: The complete structure as a free body: The sum of the moments about G: MG = +3F - 5A = 0, from which A = 3F = 6 kN which is the reaction of the floor. The 5 sum of the forces: Fy = Gy - F + A = 0, from which Gy = F - A = 10 - 6 = 4 kN. Fx = Gx = 0. Element DEG: The sum of the moments about D M = -F + 3E + 4Gy = 0, from which E = 3 The sum of the forces: F -4Gy F D A 1m E G B 1m 2m C 1m F GY GX A 2m = -2 kN. = 10-16 3 3m F GY 2m F E 1m C = -E Fy = Gy - F + E + D = 0, from which D = F - E - Gy = 10 + 2 - 4 = 8 kN. Element ABC: Noting that the reactions are equal and opposite: B = -D = -8 kN , and C = -E = 2 kN . D 1m A 1m B= D 8 kN The sum of the forces: Fy = A + B + C = 0, from which A = 8 - 2 = 6 kN. Check 2 kN C 3m A 1m 6 kN B Problem 6.56 Consider the frame in Problem 6.55. The cable CE will safely support a tension of 10 kN. Based on this criterion, what is the largest downward force F that can be applied to the frame? Solution: y From the solution to Problem 6.55: E = , 3 3 F Gy = F - A, and A = 5 F . Back substituting, E = - 5 or F = -5E, from which, for E = 10 kN, F = -50 kN F -4G Problem 6.57 The hydraulic actuator BD exerts a 6kN force on member ABC. The force is parallel to BD, and the actuator is in compression. Determine the forces on member ABC, presenting your answers as shown in Fig. 6.35. A B C 0.5 m 0.5 m D 0.5 m Solution: The surface at C is smooth. Element ABC: The sum of the moments about A is M = (0.5)B sin 45 + (1)C = 0, from which C = -3(0.707) = -2.121 kN . The sum of the forces: Fy = Ay + B sin 45 + C = 0, from which Ay = -B (0.707) - C = -2.121 kN . Fx = Ax - B cos 45 = 0, from which Ax = 4.24 kN A 0.5 m B C 0.5 m D 0.5 m C AY AX 0.5 m B 45 0.5 m 2.12 kN 4.24 kN A B 45 2.12 kN 6 kN C Problem 6.58 The simple hydraulic jack shown in Problem 6.57 is designed to exert a vertical force at point C. The hydraulic actuator BD exerts a force on the beam ABC that is parallel to BD. The largest lifting force the jack can exert is limited by the pin support A, which will safely support a force of magnitude 20 kN. What is the largest lifting force the jack can exert at C, and what is the resulting axial force in the hydraulic actuator? Solution: Ay From the solution to Problem 6.?? B Ax = , 2 B = - - C, 2 B and C = - . 2 2 Substituting, Ay = - and |A| = 2 B , 2 B A2 + A2 = x y 2 2(20) 5 B 2 12 + 1 2 2 = 5B . 2 2 For |A| = 20 kN, B= and 2 = 25.3 kN is the largest axial force, C=- 2 = -8.944 kN is the largest lifting force. Problem 6.59 Determine the forces on member BC and the axial force in member AC. B 0.3 m 0.5 m 800 N 0.4 m A C Solution: Element BC: The sum of the moments about B: M = -(0.3)800 + (0.8)C = 0, from which C = 300 N . The sum of the forces Fy = B - 800 + C = 0, from which B = 500 N . Fx = Cx = 0. (The roller support prevents an x-direction reaction in C.) Element AC: The sum of the forces Fx = Ax = 0 300 m B 500 m 800 N 400 m A C 800 N B C 300 mm Ax 500 mm C Problem 6.60 An athlete works out with a squat thrust machine. To rotate the bar ABD, he must exert a vertical force at A that causes the magnitude of the axial force in the two-force member BC to be 1800 N. When the bar ABD is on the verge of rotating, what are the reactions on the vertical bar CDE at D and E? 0.6 m 0.6 m C 0.42 m A B D 1.65 m E Solution: Member BC is a two force member. The force in BC is along the line from B to C. C y FBC 0.6 m 0.6 m C Dy 0.42 m D Dx x (FBC = 1800 N) A B D 0.42 m Ay 0.6 m 0.6 m (FBC = 1800 N) tan = 0.42 = 34.99 . 0.6 Fx : Fy : + MD : Dx - FBC cos = 0 Ay - FBC sin + Dy = 0 -1.2Ay + 0.6FBC sin = 0 1.65 m E Solving, we get Dx = 1475 N Dy = 516 N Ay = 516 N Problem 6.61 The frame supports a 6-kN load at C. Determine the reactions on the frame at A and D. A 6 kN 0.4 m B 1.0 m C 0.5 m D 0.8 m E F 0.4 m Solution: Ay Note that members BE and CF are two force members. Consider the 6 kN load as being applied to member ABC. 0.4 m A B 1.0 m 6 kN C B Ax 0.4 m 0.5 0.4 0.5 0.2 1.0 m FBE 6 kN C FCF 0.5 m D 0.8 m E F 0.4 m tan = tan = = 51.34 = 68.20 Member DEF FBE FCF F 0.4 m Dx 0.8 m Dy Equations of equilibrium: Member ABC: Fx : Fy : + MA : Fx : Fy : + MD : E Ax + FBE cos - FCF cos = 0 Ay - FBE sin - FCF sin - 6 = 0 -(0.4)FBE sin - (1.4)FCF sin - 1.4(6) = 0 Dx - FBE cos + FCF cos = 0 Dy + FBE sin + FCF sin = 0 (0.8)(FBE sin ) + 1.2FCF sin = 0 Member DEF : Unknowns Ax , Ay , Dx , Dy , FBE , FCF we have 6 eqns in 6 unknowns. Ax = -16.8 kN Ay = 11.25 kN Dx = 16.3 kN Dy = -5.25 kN Solving, we get Also, FBE = 20.2 kN (T ) FCF = -11.3 kN (C) Problem 6.62 The mass m = 120 kg. Determine the forces on member ABC, presenting your answers as shown in Fig. 6.35. 300 m m A B C D m E m 200 mm 200 mm Solution: The equations of equilibrium for the entire frame are FX = AX + EX = 0, FY = AY - 2mg = 0, and summing moments at A, MA = (0.3)EX - (0.2)mg - (0.4)mg = 0. Solving yields AX = -2354 N, AY = 2354 N, and EX = 2354 N. Member ABC: The equilibrium equations are FX = AX + CX = 0, FY = AY - BY + CY = 0, and MA = -(0.2)BY + (0.4)CY = 0. Ax Ay A B C 300 m D m mg Ex E 200 m m 200 m mg We have three equations in the three unknowns BY , CX , and CY . Solving, we get BY = 4708 N, CX = 2354 N, and CY = 2354 N. This gives all of the forces on member ABC. A similar analysis can be made for each of the other members in the frame. The results of solving for all of the forces in the frame is shown in the figure. AX AY BY BY DY DY CY CX CX CY EX 4708 N 2354 N C 2354 N B 2354 N 2354 N C D 1177 N 1177 N 2354 N B 2354 N A 4708 N 4708 N 4708 N E 2354 N Problem 6.63 The tension in cable BD is 500 lb. Determine the reactions at A for cases (1) and (2). G E 6 in D 6 in A 8 in B 8 in (1) C 300 lb Solution: Case (a) The complete structure as a free body: The sum of the moments about G: MG = -16(300) + 12Ax = 0, from which Ax = 400 lb . The sum of the forces: Fx = Ax + Gx = 0, from which Gx = -400 lb. Fy = Ay - 300 + Gy = 0, from which Ay = 300 - Gy . Element GE: The sum of the moments about E: ME = -16Gy = 0, from which Gy = 0, and from above Ay = 300 lb. Case (b) The complete structure as a free body: The free body diagram, except for the position of the internal pin, is the same as for case (a). The sum of the moments about G is MC = -16(300) + 12Ax = 0, from which Ax = 400 lb . Element ABC: The tension at the lower end of the cable is up and to the right, so that the moment exerted by the cable tension about point C is negative. The sum of the moments about C: MC = -8B sin - 16Ay = 0, noting that B = 500 lb and = tan-1 then Ay = -150 lb. 6 8 G E 6 in D 6 in A 8 in B 8 in (2) C 300 lb G E 6 in D G E 6 in D 6 in A 6 in C 300 lb 8 in 8 in (a) Gy A 8 in B 8 in (b) C 300 lb = 36.87 , (a) 12 in Gx Ay 16 in 300 lb Gy Gx Ey Ex Ax B Cy Cx 300 lb Ay (b) Ax 8 in 8 in Problem 6.64 Determine the forces on member ABCD, presenting your answers as shown in Fig. 6.35. E 3 ft 400 lb 3 ft A B C D 4 ft 4 ft 4 ft Solution: The complete structure as a free body: The sum of the moments about A: MA = -400(3) + 12Dy = 0, E 3 ft 400 lb 3 ft from which Dy = 100 lb . The sum of the forces: Fx = Ax + 400 = 0, from which Ax = -400 lb . Fy = Ay + Dy = 0, from which Ay = -100 lb . Element EC: The sum of the moments about C: MC = 6E cos - 3(400) = 0. The angle of cable element EB is = tan-1 6 4 = 56.31 , A 4 ft B C 4 ft 4 ft D E Ay F Cy Cx B Cy Cx Dy Ax from which the cable tension is E = 360.6 lb . The sum of the forces: Fx = -Cx + 400 - E cos = 0, from which Cx = 200 lb. Fy = -Cy - E sin = 0, from which Cy = -300 lb. Element ABCD: The tension in the cable acts on element ABCD with equal and opposite tension to the reaction on element EB, up and to the right at an angle of 56.31 , Cx = 200 lb to the right, and Cy = -300 lb downward. 360 lb 400 lb A 56.3 200 lb 100 lb D 100 lb 300 lb Problem 6.65 The mass m = 50 kg. Determine the forces on member ABCD, presenting your answers as shown in Fig. 6.35. D 1m 1m 1m E C 1m B 1m A F m Solution: The weight of the mass hanging is W = mg = 50(9.81) = 490.5 N The complete structure as a free body: The sum of the moments about A: MA = -2W + Fy = 0, from which Fy = 981 N. The sum of the forces: Fy = Ay + Fy - W = 0, from which Ay = -490.5 N, Fx = Ax + Fx = 0, from which Ax = -Fx . Element BF: The sum of the moments about F : MF = -Bx - By = 0, from which By = -Bx . The sum of the forces: Fy = By + Fy = 0, from which By = -981 N, and Bx = 981 N. Fx = Bx + Fx = 0, from which Fx = -981 N, and from above, Ax = 981 N , Element DE: The sum of the moments about D: MD = -Ey - 2W = 0, from which Ey = -981 N. The sum of the forces: Fy = -Dy - Ey - W = 0, from which Dy = 490.5 N . Fx = -Dx - Ex = 0, from which Dx = -Ex . Element CE: The sum of the moments about C: MC = Ey - Ex = 0, from which Ex = -981 N, and from above Dx = 981 N . Fy = Ey + Cy = 0, from which Cy = 981 N. Fx = Ex + Cx = 0, from which Cx = 981 N, and Element ABCD: All reactions on ABCD have been determined above. The components at B and C have the magnitudes B = C = 9812 + 9812 = 1387 N , at angles of 45 . 1m 1m D 1m C 1m B 1m A E W F Dy Dx Dx Dy Cy Ey Ex Ey W Ex By Cx Bx Fy Fx Cx Bx Ay Cy By Ax 490.5 N D C B A 45 1387 N 981 N 490.5 N 981 N 1387 N 45 Problem 6.66 ber BCD. Determine the forces on mem6 ft A 400 lb B 4 ft C 4 ft D E 8 ft Solution: The following is based on free body diagrams of the elements: The complete structure as a free body: The sum of the moments about D: MD = -(6)400 + 8Ey = 0, from which Ey = 300 lb. The sum of the forces: Fx = Dx = 0. Fy = Ey + Dy - 400 = 0, from which Dy = 100 lb. Element AB: The sum of the moments about A: MA = -8By - (6)400 = 0, from which By = -300 lb. The sum of forces: Fy = -By - Ay - 400 = 0, from which Ay = -100 lb. Fx = -Ax - Bx = 0, from which (1) Ax + Bx = 0 Element ACE: The sum of the moments about E: ME = -8Ax + 4Cx - 8Ay + 4Cy = 0, from which (2) -2Ax +Cx -2Ay +Cy = 0. The sum of the forces: Fy = Ay + Ey - Cy = 0, from which Cy = 200 lb . Fx = Ax - Cx = 0, from which (3) Ax = Cx . The three numbered equations are solved: Ax = -400 lb, Cx = 400 lb , and Bx = -400 lb . Element BCD: The reactions are now known: By = -300 lb , Bx = -400 lb , Cy = 200 lb , Dx = 0 , Dy = 100 lb , where negative sign means that the force is reversed from the direction shown on the free body diagram. 6 ft A 400 lb B 4 ft C 4 ft E D 8 ft Ay Ax Ax Ay Dy Dx Cy Cx 400 lb By Bx Cy Cx E Bx By Problem 6.67 Determine the forces on member ABC. E 6 kN 1m C D 1m A 2m B 2m 1m Solution: The frame as a whole: The equations of equilibrium are FX = AX + EX = 0, FY = AY + EY - 6000 N = 0, and, with moments about E, ME = 2AX - (5)6000 = 0. Solving for the support reactions, we get AX = 15,000 N and EX = -15,000 N. We cannot yet solve for the forces in the y direction at A and E. Member ABC: The equations of equilibrium are FX = AX - BX = 0, FY = AY - BY - CY = 0, and summing moments about A, MA = -2BY - 4CY = 0. Member BDE: The equations of equilibrium are FX = EX + DX + BX = 0, FY = EY + DY + BY = 0, and, summing moments about E, ME = (1)DY + (1)DX + (2)BY + (2)BX = 0. Member CD: The equations of equilibrium are FX = -DX = 0, FY = -DY + CY - 6000 = 0, and summing moments about D, MD = -(4)6000 + 3CY = 0. Solving these equations simultaneously gives values for all of the forces in the frame. The values are AX = 15,000 N, AY = -8,000 N, BX = 15,000 N, BY = -16,000 N, CY = 8,000 N, DX = 0, and DY = 2,000 N. EX 1m 1m AX EY E D C 6 kN B 2m AY 2m 1m EX EY E DY DX D DX D DY BX B BY BX B AX A BY AY CY C CY C 6 kN Problem 6.68 ber ABD. Determine the forces on mem8 in A 8 in B E 8 in C D 8 in 8 in 60 lb 60 lb Solution: The equations of equilibrium for the truss as a whole are FX = AX + CX = 0, FY = AY - 60 - 60 = 0, and MA = 16CX - 16(60) - 24(60) = 0. AY AX A 8 in. B 8 in. C CX D 60 lb 8 in. 8 in. 8 in. 60 lb E Solving these three equations yields AX = -150 lb, AY = 120 lb, and CX = 150 lb. Member ABD: The equilibrium equations for this member are: FX = AX - BX - DX = 0, FY = AY - BY - DY = 0, and MA = -8BY - 8DY - 8BX - 16DX = 0. FX = BX + EX = 0, FY = BY + EY - 60 - 60 = 0, and MB = -8(60) - 16(60) + 16EY = 0. FX = CX + DX - EX = 0, FY = DY - EY = 0, and MD = 8EX - 16EY = 0. AY AX BX DX BY 60 lb BY 60 lb EX EY DY DX EY EX Member BE: The equilibrium equations for this member are: BX B DY CX Member CDE: The equilibrium equations for this member are: Solving these equations, we get BX = -180 lb, BY = 30 lb, DX = 30 lb, DY = 90 lb, EX = 180 lb, and EY = 90 lb. Note that we have 12 equations in 9 unknowns. The extra equations provide a check. Problem 6.69 The mass m = 12 kg. Determine the forces on member CDE. A 200 mm B 200 mm C D 200 mm 400 mm m E 100 mm Solution: The equations of equilibrium for the entire truss are: FX = AX + CX = 0, FY = AY - mg = 0, and MA = 0.4CX - 0.7mg = 0. AX = -206.0 N, AY = 117.7 N, and Cx = 206.0 N. Member ABD: The equations are FX = AX + BX + DX + T = 0, FY = AY + BY + DY = 0, and MA = 0.2BY + 0.2BX + 0.2DY + 0.4DX + 0.1T = 0. FX = CX - DX + EX = 0, FY = -DY + EY = 0, and MD = 0.4EY - 0.2EX = 0. AY AX A 200 mm B 200 mm C CX D 200 mm 400 mm F m E 100 mm From these equations we get AY AX T BX DX BY DY CX DX DY mg BX BY EX EY T T PY PX T Member CDE: The equations are PY PX EY EX Member BE: The equations are FX = -BX + PX - EX = 0, FY = -BY + PY - EY = 0, and ME = 0.4BY = 0. FX = -T - PX = 0, and FY = -T - PY = 0. FY = T - mg = 0. Solving the equations simultaneously, we get BX = 117.7 N, BY = 0, DX = -29.4 N, DY = -117.7 N, EX = -235.4 N, EY = -117.7 N, T = 117.7 N, PX = -117.7 N, PY = -117.7 N The Pulley: The equations are The Weight: The equation is Problem 6.70 The weight W = 80 lb. Determine the forces on member ABCD. 11 in 5 in 12 in 3 in A 8 in B C D W E F Solution: The complete structure as a free body: The sum of the moments about A: MA = -31W + 8Ex = 0, from which Ex = 310 lb. The sum of the forces: Fx = Ex + Ax = 0, from which Ax = -310 lb . Fy = Ey + Ay - W = 0, from which (1) Ey + Ay = W . Element CFE: The sum of the forces parallel to x: Fx = Ex - Cx = 0, from which Cx = 310 lb . The sum of the moments about E: ME = 8F - 16Cy + 8Cx = 0. For frictionless pulleys, F = W , and thus Cy = 195 lb . The sum of forces parallel to y: Fy = Ey - Cy + F = 0, from which Ey = 115 lb . Equation (1) above is now solvable: Ay = -35 lb . Element ABCD: The forces exerted by the pulleys on element ABCD are, by inspection: Bx = W = 80 lb , By = 80 lb , Dx = 80 lb , and Dy = -80 lb , where the negative sign means that the force is reversed from the direction of the arrows shown on the free body diagram. 11 in. 5 in. 12 in. 3 in. A 8 in. B C D W E F Ay Ax Ey F Ex By Bx Cy Cx Cx Cy Dy Dx W Problem 6.71 The man using the exercise machine is holding the 80-lb weight stationary in the position shown. What are the reactions at the built-in support E and the pin support F ? (A and C are pinned connections.) 2 ft 2 in B 9 in A 2 ft C 1 ft 6 in D 60 6 ft 80 lb E F Solution: The complete structure as a free body: The sum of the moments about E: M = -26W - 68W sin 60 + 50Fy - 81W cos 60 + ME = 0 from which (1) 50Fy + ME = 10031. The sum of the forces: Fx = Fx + W cos 60 + Ex = 0, from which (2) Fx + Ex = -40. Fy = -W - W sin 60 + Ey + Fy = 0, from which (3) Ey + Fy = 149.28 Element CF: The sum of the moments about F : M = -72Cx = 0, from which Cx = 0. The sum of the forces: Fx = Cx + Fx = 0, from which Fx = 0 . From (2) above, Ex = -40 lb Element AE: The sum of the moments about E: M = ME - 72Ax = 0, . from which (4) ME = 72Ax . The sum of the forces: Fy = Ey + Ay = 0, from which (5) Ey + Ay = 0. Fx = Ax + Ex = 0; from which Ax = ME = 2880 in lb = 240 ft lb . the 5 reactions on E and F . 40 lb, and from (4) From (1) Fy = 143.0 lb , 2 ft 2 in B 9 in A 2 ft C 1 ft 6 in D 60 6 ft 80 lb E F 26 in 42 in 60 W W 81 in ME Ey Ex 50 in Fy Fx and from (2) Ey = 6.258 lb . This completes the determination of Cy Cx Ay Ax ME Fx Fy Ey Ex Problem 6.72 The frame supports a horizontal load F at C. The resulting compressive axial force in the twoforce member CD is 2400 N. Determine the magnitude of the reaction exerted on member ABC at B. B C F 100 mm Solution: First, write eqns to determine the support reactions at A and E (CD is a two force member) C F A B 0.3 m D AX 0.4 m AY 200 mm E EY 100 mm 100 mm E D 100 mm 100 mm C F 100 mm Fx : Fy : + MA : Ax + F = 0 Ay + Ey = 0 (1) (2) B 100 mm D 100 mm A E 0.4Ey - 0.3F = 0 (3) We have these eqns in four Unknowns (Ax , Ay , Ey , and F ) Now write eqns for member ABC CY F BY 0.1 m BX AX 0.2 m BX DY AY 0.2 m 0.1 m 0.1 m EY 0.1 m CY = 2400 N BY 200 mm 100 mm 100 mm Cy = 2400 N Fx : Fy : + MA : Ax + Bx + F = 0 Ay + By + Cy = 0 (4) (5) Cy = -Dy 0.2By - 0.2Bx + 0.3Cy - 0.3F = 0 (6) We now have 6 eqns in unknowns: (Ax , Ay , Bx , By Ey , F ) Next, write the equations for member BDE. Solving, we get Bx = 0 By = -1200 N |B| = 1200 N (two force member) Fx : Fy : MB : -Bx = 0 (7) -By + Dy + Ey = 0 (8) 0.1Dy + 0.2Ey = 0 (9) We now have 9 equations in 8 unknowns. Obviously, if they are compatible, one is a linear combination of the others. We could also have more than one redundant equation and still need another equation. Combining Eqs. (4) and (7) gives Eq. (1). Thus, one of these three equations is not need. Problem 6.73 The two-force member CD of the frame shown in Problem 6.72 will safely support a compressive axial load of 3 kN. Based on this criterion, what is the largest safe magnitude of the horizontal load F ? Solution: In the solution to Problem 6.72, we derived the equation's listed below for the loads shown on the frame. Ax + F = 0 Ay + Ey = 0 0.4Ey - 0.3F = 0 Ax + Bx + F = 0 Ay + By + Cy = 0 0.2By - 0.2Bx + 0.3Cy - 0.3F = 0 Cy = -Dy - 2 force member Set Cy = 3000 N and solve. Member ABC Entire Frame C F 100 mm B 100 mm D 100 mm A E 200 mm We get F = 2000 N = 2 kN Ax = -2 kN Ay = -1.5 kN Ey = 1.5 kN Bx = 0 By = -1.5 kN 100 mm 100 mm Problem 6.74 The unstretched length of the string is LO . Show that when the system is in equilibrium the angle satisfies the relation sin = 2(LO - 2F/k)L. 1L 4 F 1L 4 1L 2 k Solution: Since the action lines of the force F and the reaction E are co-parallel and coincident, the moment on the system is zero, and the system is always in equilibrium, for a non-zero force F . The object is to find an expression for the angle for any non-zero force F . The complete structure as a free body: The sum of the moments about A MA = -F L sin + EL sin = 0, from which E = F . The sum of forces: Fx = Ax = 0, from which Ax = 0. Fy = Ay + E - F = 0, from which Ay = 0, which completes a demonstration that F does not exert a moment on the system. The spring C: The elongation of the spring is s = 2 L sin - LO , from which the force in the spring is 4 T =k L sin - LO 2 F 1L - 4 1L - 4 C 1L - 2 k B D A E F L Element BE: The strategy is to determine Cy , which is the spring force on BE. The moment about E is ME = - from which L L L Cy cos - By cos - Bx cos = 0, 4 2 2 + By = -Bx . The sum of forces: Ax Ay By Bx L 4 E Cy 2 Fx = Bx = 0, from which Bx = 0. Fy = Cy + By + E = 0, from which Cy +By = -E = -F . The two simultaneous equations are solved: Cy = -2F , and By = F . The solution for angle : The spring force is Cy = T = k from which k Solve: sin = L sin - LO 2 L sin - LO 2 2(LO - 2F ) k L Cy E L 4 , = -2F . Problem 6.75 The pin support B will safely support a force of 24-kN magnitude. Based on this criterion, what is the largest mass m that the frame will safely support? 500 mm C 100 mm 300 mm E B D A m F 300 mm 400 mm 400 mm Solution: The weight is given by W = mg = 9.81 g The complete structure as a free body: Sum the forces in the x-direction: Fx = Ax = 0, from which Ax = 0 Element ABC: The sum of the moments about A: MA = +0.3Bx + 0.9Cx - 0.4W = 0, from which (1) 0.3Bx + 0.9Cx = 0.4W . The sum of the forces: Fx = -Bx - Cx + W + Ax = 0, from which (2) Bx + Cx = W . Solve the simultaneous equations (1) and (2) to obtain Bx = 5 W 6 Element BE: The sum of the moments about E: ME = 0.4W - 0.7By = 0, from which By = |B| = W 5 6 4 W. 7 2 C 500 mm 100 mm 300 mm A B D W 300 mm 400 mm 400 mm Cy Cx W Bx By By Bx Ay Ax W W Ex Ey Ey Ex F Cy Cx E F The magnitude of the reaction at B is 4 7 2 + = 1.0104W. 24 1.0104 = 23.752 kN is the For a safe value of |B| = 24 kN, W = maximum load that can be carried. Thus, the largest mass that can be supported is m = W/g = 23752 N/9.81 m/s2 = 2421 kg. Problem 6.76 Determine the reactions at A and C. A C 3 ft 36 lb 3 ft B 18 lb 4 ft 8 ft 72 ft-lb Solution: The complete structure as a free body: The sum of the moments about A: MA = -4(18) + 3(36) + 12Cy - 72 = 0, from which Cy = 3 lb. The sum of the forces: Fy = Ay + Cy - 18 = 0, from which Ay = 15 lb. Fx = Ax + Cx + 36 = 0, from which (1) Cx = -Ax - 36 Element AB: The sum of the forces: Fy = Ay - By - 18 = 0, from which By = -3 lb. The sum of the moments: MA = 6Bx - 4(18) - 4By - 72 = 0, from which Bx = 22 lb. The sum of the forces: Fx = Ax + Bx = 0, from which Ax = -22 lb From equation (1) Cx = -14 lb A C 3 ft 72 ft-lb 36 lb B 18 lb 4 ft 8 ft 3 ft Ay Ax 72 ft-lb Cy Cx 3 ft 36 lb 18 lb 4 ft Ay Ax 72 ft-lb 8 ft By 6 ft Bx 18 lb Problem 6.77 Determine the forces on member AD. 200 N D 400 mm 400 N 130 mm A 400 mm C B 400 mm Solution: Denote the reactions of the support by Rx and Ry . The complete structure as a free body: Fx = Rx - 400 = 0, 200 N D 400 mm 130 mm from which Rx = 400 N. The sum of moments: MA = 800C - 400(800) - 400(400) - 400(200) = 0, from which C = 300 N. Fy = C + Ry - 400 - 200 = 0, from which Ry = 300 N. Element ABC: The sum of the moments: MA = -4By + 8C = 0, from which By = 600 N. Element BD: The sum of the forces: Fy = By - Dy - 400 = 0, from which Dy = 200 N. Element AD: The sum of the forces: Fy = Ay + Dy - 200 = 0, from which Ay = 0: Element AD: The sum of the forces: Fx = Ax + Dx = 0 and MA = -400(200) + 800Dy - 400Dx = 0 A 400 mm B 400 mm 200 N Ay 400 N By Ax Rx Bx Bx By C Dy Dy Dx Dx 400 N C 400 N Ax Ry Ay Ax = -200 N, and Dx = 200 N. Element BD: The sum of forces: Fx = Bx - Dx - 400 = 0 from which Bx = 600 N. This completes the solution of the nine equations in nine unknowns, of which Ax , Ay , Dx , and Dy are the values required by the Problem. Problem 6.78 The frame shown is used to support high-tension wires. If b = 3 ft, = 30 , and W = 200 lb, what is the axial force in member HJ? C A B D E G F W H I J W W b b b b Solution: Joints B and E are sliding joints, so that the reactions are normal to AC and BF , respectively. Member HJ is supported by pins at each end, so that the reaction is an axial force. The distance h = b tan = 1.732 ft Member ABC. The sum of the forces: Fx = Ax + B sin = 0, Fy = Ay - W - B cos = 0. The sum of the moments about B: MB = bAy - hAx + bW = 0. These three equations have the solution: Ax = 173.21 lb, Ay = -100 lb, and B = -346.4 lb. Member BDEF: The sum of the forces: Fx = Dx - B sin - E sin = 0, Fy = Dy - W + B cos - E cos = 0. The sum of the moments about D: MD = -2bW - bE cos - hE sin - bB cos + hB sin = 0. These three equations have the solution: Dx = -259.8 lb, Dy = 350 lb, E = -173.2 lb. Member EGHI: The sum of the forces: Fx = Gx + E sin - H cos = 0, Fy = Gy - W + E cos + H sin = 0. The sum of the moments about H: MH = bGy - hGx + bW + 2bE cos - 2hE sin = 0. These three equations have the solution: Gx = 346.4 lb, Gy = 200 lb, and H = 300 lb. This is the axial force in HJ. A B C D E W H I G F J W W b b b b B Ay h b G y Ax B Dy Dx E Gx H W W W Problem 6.79 What are the magnitudes of the forces exerted by the pliers on the bolt at A when 30-lb forces are applied as shown? (B is a pinned connection.) 2 in 45 6 in 30 lb B A 30 lb Solution: Element AB: The sum of the moments about B: MB = 2F - (6)30 = 0, from which F = 90 lb. 45 2 in B A 6 in 30 lb 30 lb 6 in By 2 in F Bx 30 lb Problem 6.80 The weight W = 60 kip. What is the magnitude of the force the members exert on each other at D? A 3 ft B D C 2 ft 3 ft W 3 ft 3 ft Solution: Assume that a tong half will carry half the weight, and denote the vertical reaction to the weight at A by R. The complete structure as a free body: The sum of the forces: Fy = R - W = 0, from which R = W Tong-Half ACD: Element AC: The sum of the moments about A: (1) MA = 3Cy + 3Cx = 0. The sum of the forces: (2) (3) Fy = R + Cy + Ay = 0, and 2 A B C 3 ft 2 ft D 3 ft Fx = Cx + Ax = 0. Element CD: The sum of the forces: W (4) (5) Fx = Dx - P - Cx = 0, and W = 0. Fy = Dy - Cy - 2 The sum of the moments: MD = 2Cx - 3Cy - 3P + 3 W =0 2 3 ft 3 ft (6) Ay By R 2 Ax R 2 Ay Ax Cy Element AB: The sum of the forces: (7) (8) Fy = -Ay + R - By = 0, and 2 Fx = -Ax - Bx = 0. Element BD: The sum of the forces: W = 0, and Fy = By - Dy - 2 Fx = Bx - Dx + P = 0. These are ten equations in ten unknowns. These have the solution R = 60 kip. Check, Ax = -30 kip, Ay = 0, Bx = 30 kip, By = 30 kip, Cx = 30 kip, Cy = -30 kip, Dx = 110 kip, Dy = 0, and P = 80 kip. The magnitude of the force the members exert on each other at D is D = 110 kip. Bx Bx Dx P Dy Dx By Dy Cx Cx Cy (9) (10) P W 2 W 2 Problem 6.81 Figure a is a diagram of the bones and biceps muscle of a person's arm supporting a mass. Tension in the biceps muscle holds the forearm in the horizontal position, as illustrated in the simple mechanical model in Fig. b. The weight of the forearm is 9 N, and the mass m = 2 kg. (a) Determine the tension in the biceps muscle AB. (b) Determine the magnitude of the force exerted on the upper arm by the forearm at the elbow joint C. B (a) 290 mm A 50 mm m 200 mm (b) 9N 150 mm C Solution: Make a cut through AB and BC just above the elbow joint C. The angle formed by the biceps muscle with respect to the forearm is = tan-1 290 = 80.2 . The weight of the mass is 50 W = 2(9.81) = 19.62 N. The section as a free body: The sum of the moments about C is MC = -50T sin + 150(9) + 350W = 0, from which T = 166.76 N is the tension exerted by the biceps muscle AB. The sum of the forces on the section is FX = Cx + T cos = 0, from which Cx = -28.33 N. FY = Cy + T sin - 9 - W = 0, from which Cy = -135.72. The magnitude of the force exerted by the forearm on the upper arm at joint C is F = 2 2 Cx + Cy = 138.65 N (a) B 290 mm A W 200 mm C 150 mm 50 9N mm (b) T W 200 mm 9N Cy 50 150 mm mm Cx Problem 6.82 The clamp presses two blocks of wood together. Determine the magnitude of the force the members exert on each other at C if the blocks are pressed together with a force of 200 N. 50 mm 50 mm 50 mm D A 125 mm B 125 mm 125 mm E C Solution: Consider the upper jaw only. The section ABC as a free body: The sum of the moments about C is MC = 100B - 250A = 0, from which, for A = 200 N, B = 500 N. The sum of the forces: Fx = Cx - B = 0, from which Cx = 500 N, Fy = Cy + A = 0, from which Cy = -200 N. The magnitude of the reaction at C: C= 2 2 Cx + Cy = 538.52 N 125 mm B 50 mm 50 mm 50 mm D A 125 mm 125 mm E C B Cy 100 mm A Cx 250 mm Problem 6.83 The pressure force exerted on the piston is 2 kN toward the left. Determine the couple M necessary to keep the system in equilibrium. B 300 mm 45 A M C 400 mm 350 mm Solution: From the diagram, the coordinates of point B are (d, d) where d = 0.3 cos(45 ). The distance b can be determined from the Pythagorean Theorem as b = (0.35)2 - d2 . From the diagram, the angle = 37.3 . From these calculations, the coordinates of points B and C are B (0.212, 0.212), and C (0.491, 0) with all distances being measured in meters. All forces will be measured in Newtons. The unit vector from C toward B is uCB = -0.795i + 0.606j. The equations of force equilibrium at C are FX = FBC cos - 2000 = 0, and FY = N - FBC sin = 0. B 0.3 m 45 A d 0.35 m d b y FBC FBCY FBCX N c 2000 N x C Solving these equations, we get N = 1524 Newtons(N), and FBC = 2514 N. The force acting at B due to member BC is FBC uBC = -2000i + 1524j N. The position vector from A to B is rAB = 0.212i + 0.212j m, and the moment of the force acting at B about A, calculated from the cross product, is given by MF BC = 747.6k N-m (counter - clockwise). The moment M about A which is necessary to hold the system in equilibrium, is equal and opposite to the moment just calculated. Thus, M = -747.6k N-m (clockwise). y M A rAB B FBC uCB x Problem 6.84 In Problem 6.83, determine the forces on member AB at A and B. Solution: In the solution of Problem 6.83, we found that the force acting at point B of member AB was FBC uBC = -2000i + 1524j N, and that the moment acting on member BC about point A was given by M = -747.6k N-m (clockwise). Member AB must be in equilibrium, and we ensured moment equilibrium in solving Problem 6.83. From the free body diagram, the equations for force equilibrium are FX = AX + FBC uBCX = AX - 2000 N = 0, and FY = AY + FBC uBCY = AY + 1524 N = 0. FBC uCB y M AX A AY x B Thus, AX = 2000 N, and AY = -1524 N. Problem 6.85 The mechanism is used to weigh mail. A package placed at A causes the weighted point to rotate through an angle . Neglect the weights of the members except for the counterweight at B, which has a mass of 4 kg. If = 20 , what is the mass of the package at A? A 100 mm 100 mm B 30 Solution: Consider the moment about the bearing connecting the motion of the counter weight to the motion of the weighing platform. The moment arm of the weighing platform about this bearing is 100 cos(30 - ). The restoring moment of the counter weight is 100 mg sin . Thus the sum of the moments is M = 100 mB g sin - 100 mA g cos(30 - ) = 0. Define the ratio of the masses of the counter weight to the mass of the package to be RM = mB . The sum of moments equation reduces to m A A M = RM sin - cos(30 - ) = 0, = 2.8794, and the mass of the packfrom which RM = sin age is mA = R4 = 1.3892 = 1.39 kg M 100 mm cos(30-) 100 mm B 30 Problem 6.86 The scoop C of the front-end loader is supported by two identical arms, one on each side of the loader. One of the two arms (ABC) is visible in the figure. It is supported by a pin support at A and the hydraulic actuator BD. The sum of the other loads exerted on the arm, including its own weight, is F = 1.6 kN. Determine the axial force in the actuator BD and the magnitude of the reaction at A. A 0.8 m D B F 0.2 m 1m 1m C Solution: The section ABC as a free body: The sum of the moments about A: MA = 0.8BD - 2F = 0, from which BD = 4 kN. The sum of the forces: Fx = Ax + BD = 0, from which Ax = -4 kN. Fy = Ay - F = 0, from which Ay = 1.6 kN. The magnitude of the reaction at A is A= A2 x + A2 y = 4.308 kN 80 cm B D F 20 cm 1m 1m C Ay Ax 0.8 m BD F 2m Problem 6.87 The mass of the scoop is 220 kg, and its weight acts at G. Both the scoop and the hydraulic actuator BC are pinned to the horizontal member at B. The hydraulic actuator can be treated as a two-force member. Determine the forces exerted on the scoop at B and D. 0.6 m 1m D C 1m G A B 0.15 m 0.6 m 0.3 m Scoop Solution: We need to know the locations of various points in the Problem . Let us use horizontal and vertical axes and define the coordinates of point A as (0,0). All distances will be in meters (m) and all forces will be in Newtons (N). From the figure in the text, the coordinates in meters of the points in the problem are A (0, 0), B (0.6, 0), C (-0.15, 0.6), D (0.85, 1), and the x coordinate of point G is 0.9 m. The unit vector from C toward D is given by uCD = 0.928i + 0.371j, and the force acting on the scoop at D is given by D = DX i + DY j = 0.928Di + 0.371Dj. From the free body diagram of the scoop, the equilibrium equations are FX = BX + DX = 0, FY = BY + DY - mg = 0, and MB = -0.3 mg + xBD DY - yBD DX = 0. 1m D C 1m 0.6 m B A 0.15 m 0.6 m 0.3 m Scoop G From the geometry, xBD = 0.25 m, and yBD = 1 m. Solving the equations of equilibrium, we obtain BX = 719.4 N, BY = 2246 N, and D = -774.8 N (member CD is in tension). y D C B A BX BY DX DY mg x D uCD DY DX Problem 6.88 In Problem 6.87, determine the axial force in the hydraulic actuator BC. Solution: The unit vectors in the directions of the forces acting at C are uCD = 0.928i + 0.371j, uCA = 0.243i - 0.970j, and uCB = 0.781i - 0.625j. The force equilibrium equations at C are FX = TBC uCBX + TAC uCAX + TCD uCDX = 0, and FY = TBC uCBY + TAC uCAY + TCD uCDY = 0. D TAC TCD C C TCD A TBC TBC Solving these equations, we get TBC = -1267 N(compression), and TAC = 1112 N(tension). TAC TCD C TBC TAC Problem 6.89 Determine the force exerted on the bolt by the bolt cutters. 75 mm A B D 40 mm C 55 mm 100 N 90 mm 60 mm 65 mm 300 mm 100 N Solution: The equations of equilibrium for each of the members will be developed. Member AB: The equations of equilibrium are: FX = AX + BX = 0, FY = AY + BY = 0, and MB = 90F - 75AX - 425(100) = 0 FX = -BX + DX = 0, FY = -BY + DY + 100 = 0, and MB = 15DX + 60DY + 425(100) = 0. F 75 mm 100 N A B D F 75 mm 40 mm C 55 mm Member BD: The equations are 90 mm 60 mm 65 mm 300 mm 100 N 100 N Member AC: The equations are FX = -AX + CX = 0, FY = -AY + CY + F = 0, and MA = -90F + 125CY + 40CX = 0. FX = -CX - DX = 0, FY = -CY - DY = 0. Solving the equations simultaneously (we have extra (but compatible) equations, we get F = 1051 N, AX = 695 N, AY = 1586 N, BX = -695 N, BY = -435 N, CX = 695 N, CY = 535 N, DX = -695 N, and Dy = -535 N 75 mm AX AY 40 mm 55 mm A BX B BY Member CD: The equations are: 90 mm 60 mm 65 mm 300 mm AX AY CY 40 mm F D C C 55 mm X 90 mm 60 mm 65 mm 300 mm BY B BX D DY DX 60 mm 65 mm 300 mm 100 N CY C DX D DY CX Problem 6.90 For the bolt cutters in Problem 6.89, determine the magnitude of the force the members exert on each other at the pin connection B and the axial force in the two-force member CD. Solution: From the solution to 6.107, we know BX = -695 N, and BY = -435 N. We also know that CX = 695 N, and CY = 535 N, from which the axial load in member CD can be calculated. The load in CD is given by TCD = 2 2 CX + CY = 877 N 25 0 mm 25 A 0m m 25 0m B m Problem 6.91 The device is designed to exert a large force on the horizontal bar at A for a stamping operation. If the hydraulic cylinder DE exerts an axial force of 800 N and = 80 , what horizontal force is exerted on the horizontal bar at A? 90 D C 400 mm E Solution: Define the x-y coordinate system with origin at C. The projection of the point D on the coordinate system is Ry = 250 sin = 246.2 mm, and Rx = 250 cos = 43.4 mm. The angle formed by member DE with the positive x axis is = Ry 180 - tan-1 400-R = 145.38 . The components of the x force produced by DE are Fx = F cos = -658.3 N, and Fy = F sin = 454.5 N. The angle of the element AB with the positive x axis is = 180 - 90 - = 10 , and the components of the force for this member are Px = P cos and Py = P sin , where P is to be determined. The angle of the arm BC with the positive x axis is = 90 + = 170 . The projection of point B is Lx = 250 cos = -246.2 mm, and Ly = 250 sin = 43.4 mm. Sum the moments about C: MC = Rx Fy - Ry Fx + Lx Py - Ly Px = 0. Substitute and solve: P = 2126.36 N, and Px = P cos = 2094 N is the horizontal force exerted at A. 250 mm A B 90 D C 250 mm 250 mm E 400 mm Fy B Px Cy Py D Fx Cx Problem 6.92 This device raises a load W by extending the hydraulic actuator DE. The bars AD and BC are 4 ft long, and the distances b = 2.5 ft and h = 1.5 ft. If W = 300 lb, what force must the actuator exert to hold the load in equilibrium? b W A B h C D E The angle ADC is = sin-1 distance CD is d = 4 cos . Solution: h 4 = 22.02 . The b W F A B h C D E The complete structure as a free body: The sum of the forces: Fy = -W + Cy + Dy = 0. Fx = Cx + Dx = 0. The sum of the moments about C: MC = -bW + dDy = 0. These have the solution: Cy = 97.7 lb, Dy = 202.3 lb, and Cx = -Dx . Divide the system into three elements: the platform carrying the weight, the member AB, and the member BC. The Platform: (See Free body diagram) The moments about the point A: MA = -bW - dB = 0. The sum of the forces: Fy = A + B + W = 0. These have the solution: B = -202.3 lb, and A = -97.7 lb. Element BC: The sum of the moments about E is MC = - (1) h 2 Cy + d 2 Cx + d 2 B = 0, from which (3) Cy - Ey + B = 0 Element AD: The sum of the moments about E: ME = from which (4) dDy + hDx - dA = 0. These are four equations in the four unknowns: EX , EY , Dx , CX and DX Solving, we obtain Dx = -742 lb. d 2 Dy + h 2 Dx - d 2 A = 0, W A Ex Cy Cx E Dx A B Ey Ex Dy B dCx - hCy - dB = 0. The sum of the forces: Fx = Cx - Ex = 0, from which (2) Ex - Cx = 0, Fy = Cy - Ey + B = 0, from which Problem 6.93 The linkage is in equilibrium under the action of the couples MA and MB . If A = 60 and B = 70 , what is the ratio MA /MB ? 250 mm MB B MA A 150 mm 350 mm Solution: Make a cut through the linkage connecting the two cranks, and treat each system as a free body. The equilibrium condition occurs when the reaction forces in the linkage are equal and opposite. The position vector of the end of the system B crank is rB = RB (i cos B + i sin B ) = 85.51i + 234.92j (mm). The position vector at the end of the system A crank is rA = RA (i cos A + j sin A ) = 75i + 129.9j (mm). The angle of the linkage from the end of the system B crank with respect to the horizontal is = tan-1 yA - yB xA - xB + 350 = -17.19 . 200 mm MB B MA A 150 mm 350 mm The unit vector parallel to the linkage, originating at the B crank, is eBA = i cos + j sin = 0.9553i - 0.2955j. The unit vector originating at A crank is eAB = -eBA . The components of the forces in the linkage are |F|eAB , and |F|eBA . System B: When the system is in equilibrium, MB + |F| 0 85.5 0.9553 0 234.9 -0.2955 1 0 = 0, 0 |F| MB MA |F| from which MB = 249.7|F|. System A: When the system is in equilibrium: MA + |F| 0 75 -0.9553 0 129.9 0.2955 1 0 = 0, 0 from which MA = -146.27|F|. Complete system: Both systems are in equilibrium for the value |F|. Take the ratio of the two moments to eliminate |F|. MA 146.27 =- = -0.5858 MB 249.7 Problem 6.94 A load W = 2 kN is supported by the members ACG and the hydraulic actuator BC. Determine the reactions at A and the compressive axial force in the actuator BC. A 0.75 m B 1m G 0.5 m W 1.5 m 1.5 m C Solution: The sum of the moments about A is MA = 0.75BC - 3(2) = 0, from which BC = 8 kN is the axial force. The sum of the forces FX = AX + BC = 0, from which AX = -8 kN. FY = AY - 2 = 0, from which AY = 2 kN. A 0.75 m B 1.0 m G 0.5 m W 1.5 m 1.5 m C AY 0.75 m AX BC W 3m Problem 6.95 The dimensions are a = 260 mm, b = 300 mm, c = 200 mm, d = 150 mm, e = 300 mm, and f = 520 mm. The ground exerts a vertical force F = 7000 N on the shovel. The mass of the shovel is 90 kg and its weight acts at G. The weights of the links AB and AD are negligible. Determine the horizontal force P exerted at A by the hydraulic piston and the reactions on the shovel at C. Solution: The free-body diagram of the shovel is from which we obtain the equations Fx = Cx - T cos = 0, (1) a B P A b Shovel D d C G Fy = Cy + T sin + F - mg = 0, (2) M(ptC) = f F - emg + (b - c)T sin +dT cos = 0. The angle = arctan[(a - d)/b]. From the free-body diagram of joint A, B P T (3) c e f F T we obtain the equation F = P + T cos = 0. (4) Substituting the given information into Eqs. (1)(4) and solving, we obtain T = -19, 260 N, P = 18, 080 N, Cx = -18, 080 N, and Cy = 513 N. d b-c CX CY mg e f F Problem 6.96 The truss supports a load F = 10 kN. Determine the axial forces in the members AB, AC, and BC. B 3m A C D 4m F 3m Solution: Fx : Fy : + MA : Find the support reactions at A and D. Ax = 0 Ay + Dy - 10 = 0 (-4)(10) + 7Dy = 0 B 3m A C D Solving, Ax = 0, Ay = 4.29 kN Dy = 5.71 kN Joint A: tan = 3 4 4m F 3m = 36.87 (Ay = 4.29 kN) Fx : Fy : Solving, FAB cos + FAC = 0 Ay + FAB sin = 0 FAB = -7.14 kN (C) FAC = 5.71 kN (T ) Joint C: Fx : Fy : FCD - FAC = 0 FBC - 10 kN = 0 AY 10 kN AX 4m 3m DY 3m FAB y Solving FBC = 10 kN (T ) FCD = +5.71 kN (T ) FAC AY x FBC FAC FCD 10 kN Problem 6.97 Each member of the truss shown in Problem 6.96 will safely support a tensile force of 40 kN and a compressive force of 32 kN. Based on this criterion, what is the largest downward load F that can safely be applied at C? A B 1 C 4 5 2 3m D 3 4m F 3m Solution: Assume a unit load F and find the magnitudes of the tensile and compressive loads in the truss. Then scale the load F up (along with the other loads) until either the tensile limit or the compressive limit is reached. External Support Loads: Fx : Fy : MA : Joint A: tan = 3 2 Ax = 0 (1) Ay + Dy - F = 0 (2) -4F + 7Dy = 0 (3) AX F AY y FAB DY = 36.87 Fx : Fy : Joint C Fx : Fy : Joint D tan = 3 3 FCD - FAC = 0 (6) FBC - F = 0 (7) FAC + FAB cos = 0 (4) FAB sin + Ay = 0 (5) x FAC AY y FBD = 45 Fx : Fy : -FCD - FBD cos = 0 (8) FBD sin + Dy = 0 (9) x FCD DY Setting F = 1 and solving, we get the largest tensile load of 0.571 in AC and CD. The largest compressive load is 0.808 in member BD. Largest Tensile is in member BC. BC = F = 1 The compressive load will be the limit Fmax 32 = 1 0.808 Fmax = 40 kN y FBC FAC F FCD x Problem 6.98 The Pratt bridge truss supports loads at F , G, and H. Determine the axial forces in members BC, BG, and F G. Solution: The angles of the cross-members are = 45 . The complete structure as a free body: The sum of the moments about A: MA = -60(4) - 80(8) - 20(12) + 16E = 0, from which E = 70 kN. The sum of the forces: Fx = Ax = 0. Fy = Ay - 60 - 80 - 20 + E = 0, from which Ay = 90 kN The method of joints: Joint A: FY = Ay + AB sin = 0, from which AB = -127.3 kN (C), Fx = AB cos + AF = 0, from which AF = 90 kN (T ). Joint F: Fx = -AF + F G = 0, B C D 4m A F 60 kN 4m 4m G 80 kN 4m H 20 kN 4m E B C D 4m A F 60 kN G 80 kN H 20 kN E 4m from which F G = 90 kN (T ) . Fy = BF - 60 = 0, from which BF = 60 kN (C). Joint B: Fx = -AB cos + BC + BG cos = 0, and Fy = -AB sin - BF - BG sin = 0, 4m 4m 4m 4m Ax Ay AB AF 4m 4m 4m 4m from which: -AB sin - BF - BG sin = 0. Solve: BG = 42.43 kN (T ) , and - AB cos + BC + BG cos = 0, from which BC = -120 kN (C) 60 kN 80 kN 20 kN BF AF FG 60 kN Joint F E Ay Joint A BC AB BF BG Joint B Problem 6.99 Consider the truss in Problem 6.98. Determine the axial forces in members CD, GD, and GH. CG BC CG CD BG GD GH Joint C 80 kN Joint G Solution: Use the results of the solution of Problem 6.98: BC = -120 kN (C), BG = 42.43 kN (T ), and F G = 90 kN (T ). The angle of the cross-members with the horizontal is = 45 . Joint C: Fx = -BC + CD = 0, from which CD = -120 kN (C) FY = -CG = 0, from which CG = 0. Joint G: Fy = BG sin + GD sin + CG - 80 = 0, from which GD = 70.71 kN (T ) . Fy = -BG cos + GD cos - F G + GH = 0, from which GH = 70 kN (T ) Problem 6.100 The truss supports a 400-N load at G. Determine the axial forces in members AC, CD, and CF 400 N A C E G 300 mm 600 mm F D B 300 mm 300 mm 300 mm H Solution: The complete structure as a free body: The sum of the moments about A: MA = -900(400) + 600B = 0, from which B = 600 N. The sum of forces: Fx = Ax + B = 0, from which Ax = -600 N. Fy = Ay - 400 = 0, from which Ay = 400 N. The method of joints: The angle from the horizontal of element BD is = tan-1 300 900 = 18.43 . 400 N A C E G 300 mm 600 mm F D B 300 mm 300 mm 900 mm 400 N 300 mm H Ay Ax 600 mm B The angle from the horizontal of element AD is AD = 90 - tan -1 300 600 - 300 tan 300 600(1 - tan ) = 59.04 . AB B Joint B CD AD AD DF BD Joint D The angle from the horizontal of element CF is CF = 90 - tan-1 Joint B: Fx = B + BD cos = 0, from which BD = -632.5 N (C) Fy = AB + BD sin = 0, from which AB = 200 N (T ) Joint A: Fy = Ay - AD sin AD - AB = 0, from which AD = 233.2 N (T ) Fx = Ax + AC + AD cos AD = 0, from which AC = 480 N (T ) = 53.13 . BD AX AY AC AD AD AB Joint A AC CE CF CF CD Joint C Joint D: Fx = -AD cos AD - BD cos + DF cos = 0, from which DF = -505.96 N (C) Fy = AD sin AD + CD - BD sin + DF sin = 0, from which CD = -240 N (C) Joint C: Fy = -CD - CF sin CF = 0, from which CF = 300 N (T ) Problem 6.101 Consider the truss in Problem 6.100. Determine the axial forces in members CE, EF , and EH. Solution: Use the results of the solution of Problem 6.100: AC = 480 N (T ), CF = 300 N (T ), DF = -505.96 N (C), = 18.4 , CF = 53.1 . The method of joints: The angle from the horizontal of element EH is EH = 90 - tan-1 Joint C: Fx = -AC + CE + CF cos CF = 0, from which CE = 300 N (T ) 300 600 - 900 tan = 45 Joint F: Fy = -CF cos CF - DF cos + F H cos = 0, from which F H = -316.2 N (C) Fy = EF + CF sin CF - DF sin + F H sin = 0, from which EF = -300 N (C) Joint E: Fy = -EH sin EH - EF = 0, from which EH = 424.3 N (T ) AC CE CF CD CF CF CF DF EF FH CE EF EG EH EH Joint C Joint F Joint E Problem 6.102 The mass m = 120 kg. Determine the forces on member ABC. A B C Solution: The weight of the hanging mass is given by m s2 = 1177 N. 300 mm D m W = mg = 120 kg 9.81 The complete structure as a free body: The equilibrium equations are: FX = AX + EX = 0, FY = AY - W = 0, and MA = 0.3EX - 0.4W = 0. AX = -1570 N, AY = 1177 N, and EX = 1570 N. Element ABC: The equilibrium equations are FX = Ax + CX = 0, FY = AY + CY - BY - W = 0, and: MA = -0.2BY + 0.4cY - 0.4W = 0. E 200 mm 200 mm Solving, we get A B 300 mm D E W 200 mm 200 mm Cy Cx Solution gives BY = 2354 N (member BD is in tension), CX = 1570 N, and CY = 2354 N. Ay Ax B B B B Ex W Cx Cy Problem 6.103 Determine the forces on member ABC, presenting your answers as shown in Fig. 6.35. 2 ft D 400 lb 200 ft-lb 1 ft 1 ft A B 100 lb E C 2 ft 2 ft 2 ft Solution: moments: The complete structure as a free body: The sum of the MA = 100(1) - 400(6) - 200 + 4E = 0, from which E = 625 lb. The sum of the forces: Fy = Ay + E - 400 = 0, from which Ay = -225 lb. Fx = Ax + 100 = 0, from which Ax = -100 lb. These results are used as a check on the solution below. Element ECD: (See the free body diagram.) The sum of the moments about E: ME = -4Dx - 2Cx - 100 = 0, from which (1) 4Dx + 2Cx = -100. The sum of the forces: Fx = Dx + Cx + 100 = 0, from which (2) Dx + Cx = -100. Fy = E + Cy + Dy = 0, thus (3) Dy + Cy + E = 0. Element BD: The sum of the moments about B: MB = 2Dx - 2Dy = 0, from which (4) Dx - Dy = 0. The sum of the forces: Fx = Bx - Dx = 0, from which (5) Bx - Dx = 0. Fy = By - Dy = 0, from which (6) By - Dy = 0 Element ABC: The sum of the moments about A: MA = -2By - 4Cy - 200 - 6(400) = 0, from which (7) By + 2Cy = -1300. The sum of the forces: D 2 ft A B 100 lb E 2 ft 2 ft 2 ft Dy Dy Bx Bx Cx Dx 400 lb Cy Cx 400 lb 200 ft-lb 1 ft 1 ft C Dx By Ay Ax By Cy200 ft-lb 100 lb E 675 lb 100 lb 225 lb 50 lb 50 lb 150 lb 400 lb 200 ft lb Fx = Ax - Bx - Cx = 0, from which (8) Ax - Bx - Cx = 0. Fy = Ay - By - Cy - 400 = 0, from which (9) Ay - By - Cy = 400. These nine equations are solved for the nine reactions The reactions are DX = 50 lb, DY = 50 lb,: CX = -150 lb, CY = -675 lb, BX = 50 lb, BY = 50 lb, AX = -100 lb, AY = -225 lb, and E = 625 lb. Problem 6.104 Determine the force exerted on the bolt by the bolt cutters and the magnitude of the force the members exert on each other at the pin connection A. A 90 N 540 mm Solution: Element AB: The moment about A is MA = -10B - 54F = 0, where F = 90 N. From which B = -486 N. The sum of the forces: Fy = A + B - F = 0, from which A = 576 N Element BC: The moment about C: MC = -16B - 8FC = 0, from which the cutting force is FC = 972 N 80 mm 160 mm 100 mm 90 N C B 90 N A FC 8 cm B 16 cm 10 cm 54 cm Problem 6.105 The 600-lb weight of the scoop acts at a point 1 ft 6 in. to the right of the vertical line CE. The line ADE is horizontal. The hydraulic actuator AB can be treated as a two-force member. Determine the axial force in the hydraulic actuator AB and the forces exerted on the scoop at C and E. B C 2 ft A 5 ft D 1 ft E 2 ft 6 in Scoop 1 ft 6 in Solution: The free body diagrams are shown at the right. Place the coordinate origin at A with the x axis horizontal. The coordinates (in ft) of the points necessary to write the needed unit vectors are A (0, 0), B (6, 2), C (8.5, 1.5), and D (5, 0). The unit vectors needed for this problem are uBA = -0.949i - 0.316j, uBC = 0.981i - 0.196j, and uBD = -0.447i - 0.894j. The scoop: The equilibrium equations for the scoop are FX = -TCB uBCX + EX = 0, FY = -TCB uBCY + EY - 600 = 0, and MC = 1.5EX - 1.5(600 lb) = 0. EX = 600 lb, EY = 480 lb, and TCB = 611.9 lb. Joint B: The equilibrium equations for the scoop are FX = TBA uBAX + TBD uBDX + TCB uBCX = 0, and FY = TBA uBAY + TBD uBDY + TCB uBCY = 0. TBA = 835 lb, and TBD = -429 lb. B 2 ft A 5 ft D 1 ft E 2 ft 6 in C 1 ft 6 in 1 ft 6" 600 lb TCB 1.5 ft EX EY C 1.5 ft G E Solving, we get 600 lb y TBA TCB TBD x Solving, we get ... View Full Document

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