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Solution Manual for Mechanics of Materials SI, 9th Edition

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    1 1–1. The shaft is supported by a smooth thrust bearing at B and a journal bearing at C . Determine the resultant internal loadings acting on the cross section at E . Support Reactions: We will only need to compute C y by writing the moment equation of equilibrium about B with reference to the free-body diagram of the entire shaft, Fig. a . a Internal Loadings: Using the result for C y , section DE of the shaft will be considered. Referring to the free-body diagram, Fig. b , Ans. Ans. a Ans. The negative signs indicates that V E and M E act in the opposite sense to that shown on the free-body diagram. M E = - 2400 lb # ft = - 2.40 kip # ft 1000(4) - 800(8) - M E = 0 + © M E = 0; V E = - 200 lb V E + 1000 - 800 = 0 + c © F y = 0; N E = 0 : + © F x = 0; C y = 1000 lb C y (8) + 400(4) - 800(12) = 0 + © M B = 0; A E D B C 4 ft 400 lb 800 lb 4 ft 4 ft 4 ft Ans: , , M E = - 2.40 kip # ft V E = - 200 lb N E = 0

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    2 1–2. Determine the resultant internal normal and shear force in the member at (a) section a–a and (b) section b–b , each of which passes through point A . The 500-lb load is applied along the centroidal axis of the member. (a) Ans. Ans. (b) Ans. Ans. V b = 250 lb V b - 500 sin 30° = 0 + Q© F y = 0; N b = 433 lb N b - 500 cos 30° = 0 R + © F x = 0; V a = 0 + T© F y = 0; N a = 500 lb N a - 500 = 0 : + © F x = 0; 30  A b a b a 500 lb 500 lb Ans: , , V b = 250 lb N b = 433 lb, V a = 0 N a = 500 lb

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    3 1–3. The beam AB is fixed to the wall and has a uniform weight of 80 lb ft. If the trolley supports a load of 1500 lb, determine the resultant internal loadings acting on the cross sections through points C and D . > Segment BC: Ans. Ans. a Ans. Segment BD: Ans. Ans. a Ans. M D = - 0.360 kip # ft - M D - 0.24 (1.5) = 0 + © M D = 0; V D = 0.240 kip V D - 0.24 = 0 + c © F y = 0; N D = 0 ; + © F x = 0; M C = - 47.5 kip # ft - M C - 2(12.5) - 1.5 (15) = 0 + © M C = 0; V C = 3.50 kip V C - 2.0 - 1.5 = 0 + c© F y = 0; N C = 0 ; + © F x = 0; D 5 ft 20 ft 3 ft 10 ft C B A 1500 lb Ans: , , , , M D = - 0.360 kip # ft V D = 0.240 kip N D = 0 M C = - 47.5 kip # ft, V C = 3.50 kip N C = 0

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    4 *1–4. The shaft is supported by a smooth thrust bearing at A and a smooth journal bearing at B . Determine the resultant internal loadings acting on the cross section at C . Support Reactions: We will only need to compute B y by writing the moment equation of equilibrium about A with reference to the free-body diagram of the entire shaft, Fig. a. a Internal Loadings: Using the result of B y , section CD of the shaft will be considered. Referring to the free-body diagram of this part, Fig. b , Ans. Ans. a Ans. The negative sign indicates that V C act in the opposite sense to that shown on the free-body diagram. M C = 433 N # m 1733.33(2.5) - 600(1)(0.5) - 900(4) - M C = 0 + © M C = 0; V C = - 233 N V C - 600(1) + 1733.33 - 900 = 0 + c© F y = 0; N C = 0 ; + © F x = 0; B y = 1733.33 N B y (4.5) - 600(2)(2) - 900(6) = 0 + © M A = 0; A D B C 900 N 1.5 m 600 N / m 1.5 m 1 m 1 m 1 m

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    5 1–5. Determine the resultant internal loadings in the beam at cross sections through points D and E . Point E is just to the right of the 3-kip load. Support Reactions: For member AB a Equations of Equilibrium: For point D Ans. Ans. a Ans. Equations of Equilibrium: For point E Ans. Ans. a Ans. Negative signs indicate that M E and V E act in the opposite direction to that shown on FBD. M E = - 24.0 kip # ft + © M E = 0; M E + 6.00(4) = 0 V E = - 9.00 kip + c © F y = 0; - 6.00 - 3 - V E = 0 : + © F x = 0; N E = 0 M D = 13.5 kip # ft + © M D = 0; M D + 2.25(2) - 3.00(6) = 0 V D = 0.750 kip + c © F y = 0; 3.00 - 2.25 - V D = 0 : + © F x = 0; N D = 0 + c © F y = 0; B y + 3.00 - 9.00 = 0 B y = 6.00 kip : + © F x = 0; B x = 0 + © M B = 0; 9.00(4) - A y (12) = 0 A y = 3.00 kip 6 ft 4 ft A 4 ft B C D E 6 ft 3 kip 1.5 kip / ft Ans: , , , , M E = - 24.0 kip # ft V E = - 9.00 kip N E = 0 M D = 13.5 kip # ft, V D = 0.750 kip N D = 0

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    6 1–6. Determine the normal force, shear force, and moment at a section through point C . Take P = 8 kN. Support Reactions: a Equations of Equilibrium: For point C Ans. Ans. a Ans. Negative signs indicate that N C and V C act in the opposite direction to that shown on FBD. M C = 6.00 kN # m + © M C = 0; 8.00(0.75) - M C = 0 V C = - 8.00 kN + c © F y = 0; V C + 8.00 = 0 N C = - 30.0 kN : + © F x = 0; - N C - 30.0 = 0 + c © F y = 0; A y - 8 = 0 A y = 8.00 kN : + © F x = 0; 30.0 - A x = 0 A x = 30.0 kN + © M A = 0; 8(2.25) - T (0.6) = 0 T = 30.0 kN 0.75 m C P A B 0.5 m 0.1 m 0.75 m 0.75 m Ans: M C = 6.00 kN # m V C = - 8.00 kN, N C = - 30.0 kN,

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    7 Support Reactions: a Ans. Equations of Equilibrium: For point C Ans. Ans. a Ans. Negative signs indicate that N C and V C act in the opposite direction to that shown on FBD. M C = 0.400 kN # m + © M C = 0; 0.5333(0.75) - M C = 0 V C = - 0.533 kN + c © F y = 0; V C + 0.5333 = 0 N C = - 2.00 kN : + © F x = 0; - N C - 2.00 = 0 + c © F y = 0; A y - 0.5333 = 0 A y = 0.5333 kN : + © F x = 0; 2 - A x = 0 A x = 2.00 kN P = 0.5333 kN = 0.533 kN + © M A = 0; P (2.25) - 2(0.6) = 0 1–7. The cable will fail when subjected to a tension of 2 kN. Determine the largest vertical load P the frame will support and calculate the internal normal force, shear force, and moment at the cross section through point C for this loading. 0.75 m C P A B 0.5 m 0.1 m 0.75 m 0.75 m Ans: , , , M C = 0.400 kN # m V C = - 0.533 kN N C = - 2.00 kN P = 0.533 kN

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    8 Referring to the FBD of the entire beam, Fig. a , a Referring to the FBD of this segment, Fig. b , Ans. Ans. a Ans. + © M C = 0; M C + 6(0.5) - 7.5(1) = 0 M C = 4.50 kN # m + c © F y = 0; 7.50 - 6 - V C = 0 V C = 1.50 kN : + © F x = 0; N C = 0 + © M B = 0; - A y (4) + 6(3.5) + 1 2 (3)(3)(2) = 0 A y = 7.50 kN *1–8. Determine the resultant internal loadings on the cross section through point C . Assume the reactions at the supports A and B are vertical. 0.5 m 0.5 m 1.5 m 1.5 m C A B 3 kN / m 6 kN D

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    9 Referring to the FBD of the entire beam, Fig. a , a Referring to the FBD of this segment, Fig. b , Ans. Ans. a Ans. = 3.94 kN # m + © M D = 0; 3.00(1.5) - 1 2 (1.5)(1.5)(0.5) - M D = 0 M D = 3.9375 kN # m + c © F y = 0; V D - 1 2 (1.5)(1.5) + 3.00 = 0 V D = - 1.875 kN : + © F x = 0; N D = 0 + © M A = 0; B y (4) - 6(0.5) - 1 2 (3)(3)(2) = 0 B y = 3.00 kN 1–9. Determine the resultant internal loadings on the cross section through point D . Assume the reactions at the supports A and B are vertical. Ans: M D = 3.94 kN # m N D = 0, V D = - 1.875 kN, 0.5 m 0.5 m 1.5 m 1.5 m C A B 3 kN / m 6 kN D

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    10 Equations of Equilibrium: For point A Ans. Ans. a Ans. Negative sign indicates that M A acts in the opposite direction to that shown on FBD. Equations of Equilibrium: For point B Ans. Ans. a Ans. Negative sign indicates that M B acts in the opposite direction to that shown on FBD. Equations of Equilibrium: For point C Ans. Ans. a Ans. Negative signs indicate that N C and M C act in the opposite direction to that shown on FBD. M C = - 8125 lb # ft = - 8.125 kip # ft + © M C = 0; - M C - 650(6.5) - 300(13) = 0 N C = - 1200 lb = - 1.20 kip + c © F y = 0; - N C - 250 - 650 - 300 = 0 ; + © F x = 0; V C = 0 M B = - 6325 lb # ft = - 6.325 kip # ft + © M B = 0; - M B - 550(5.5) - 300(11) = 0 V B = 850 lb + c © F y = 0; V B - 550 - 300 = 0 ; + © F x = 0; N B = 0 M A = - 1125 lb # ft = - 1.125 kip # ft + © M A = 0; - M A - 150(1.5) - 300(3) = 0 V A = 450 lb + c © F y = 0; V A - 150 - 300 = 0 ; + © F x = 0; N A = 0 1–10. The boom DF of the jib crane and the column DE have a uniform weight of 50 lb ft. If the hoist and load weigh 300 lb, determine the resultant internal loadings in the crane on cross sections through points A , B , and C . > 5 ft 7 ft C D F E B A 300 lb 2 ft 8 ft 3 ft Ans: , , , , , M C = - 8.125 kip # ft N C = - 1.20 kip, V C = 0 M B = - 6.325 kip # ft, V B = 850 lb N B = 0 M A = - 1.125 kip # ft, V A = 450 lb N A = 0

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    11 1–11. The forearm and biceps support the 2-kg load at A . If C can be assumed as a pin support, determine the resultant internal loadings acting on the cross section of the bone of the forearm at E .The biceps pulls on the bone along BD . Support Reactions: In this case, all the support reactions will be completed. Referring to the free-body diagram of the forearm, Fig. a , a Internal Loadings: Using the results of C x and C y , section CE of the forearm will be considered. Referring to the free-body diagram of this part shown in Fig. b , Ans. Ans. a Ans. The negative signs indicate that N E , V E and M E act in the opposite sense to that shown on the free-body diagram. M E = - 2.26 N # m M E + 64.47(0.035) = 0 + © M E = 0; V E = - 64.5 N - V E - 64.47 = 0 + c© F y = 0; N E = - 22.5 N N E + 22.53 = 0 : + © F x = 0; C y = 64.47 N 87.05 sin 75° - 2(9.81) - C y = 0 + c © F y = 0; C x = 22.53 N C x - 87.05 cos 75° = 0 : + © F x = 0; F BD = 87.05 N F BD sin 75°(0.07) - 2(9.81)(0.3) = 0 + © M C = 0; 75  230 mm 35 mm 35 mm C E B D A Ans: , , M E = - 2.26 N # m V E = - 64.5 N N E = - 22.5 N

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    12 *1–12. The serving tray T used on an airplane is supported on each side by an arm. The tray is pin connected to the arm at A , and at B there is a smooth pin. (The pin can move within the slot in the arms to permit folding the tray against the front passenger seat when not in use.) Determine the resultant internal loadings acting on the cross section of the arm through point C when the tray arm supports the loads shown. Ans. Ans. a Ans. M C = - 9.46 N # m + © M C = 0; - M C - 9(0.5 cos 60° + 0.115) - 12(0.5 cos 60° + 0.265) = 0 V C = 10.5 N V C - 9 sin 30° - 12 sin 30° = 0; a + © F y = 0; N C = - 18.2 N N C + 9 cos 30° + 12 cos 30° = 0; b+ © F x = 0; 9 N 500 mm 12 N 15 mm 150 mm 60  A B C T V C M C N C 100 mm

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    13 Internal Loadings: Referring to the free-body diagram of the section of the hacksaw shown in Fig. a , Ans. Ans. a Ans. The negative sign indicates that N a–a and M a–a act in the opposite sense to that shown on the free-body diagram. M a - a = - 15 N # m - M a - a - 100(0.15) = 0 + © M D = 0; V a - a = 0 + c © F y = 0; N a - a = - 100 N N a - a + 100 = 0 ; + © F x = 0; 1–13. The blade of the hacksaw is subjected to a pretension force of Determine the resultant internal loadings acting on section a–a that passes through point D . F = 100 N. A B C D F F a b b a 30  225 mm 150 mm Ans: , , M a - a = - 15 N # m V a - a = 0 N a - a = - 100 N

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    14 1–14. The blade of the hacksaw is subjected to a pretension force of . Determine the resultant internal loadings acting on section b–b that passes through point D . F = 100 N Internal Loadings: Referring to the free-body diagram of the section of the hacksaw shown in Fig. a , Ans. Ans. a Ans. The negative sign indicates that N b–b and M b–b act in the opposite sense to that shown on the free-body diagram. M b - b = - 15 N # m - M b - b - 100(0.15) = 0 + © M D = 0; V b - b = 50 N © F y ¿ = 0; V b - b - 100 sin 30° = 0 N b - b = - 86.6 N © F x ¿ = 0; N b - b + 100 cos 30° = 0 A B C D F F a b b a 30  225 mm 150 mm Ans: , , M b - b = - 15 N # m V b - b = 50 N N b - b = - 86.6 N

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    15 1–15. A 150-lb bucket is suspended from a cable on the wooden frame. Determine the resultant internal loadings on the cross section at D . Support Reactions: We will only need to compute B x , B y , and F GH . Referring to the free-body diagram of member BC, Fig. a, a Internal Loadings: Using the results of B x and B y , section BD of member BC will be considered. Referring to the free-body diagram of this part shown in Fig. b , Ans. Ans. a Ans. The negative signs indicates that V D and M D act in the opposite sense to that shown on the free-body diagram. M D = - 150 lb # ft 150(1) + M D = 0 + © M D = 0; V D = - 150 lb - V D - 150 = 0 + c © F y = 0; N D = 300 lb N D - 300 = 0 : + © F x = 0; B y = 150 lb 424.26 sin 45° - 150 - B y = 0 + c© F y = 0; B x = 300 lb 424.26 cos 45° - B x = 0 : + © F x = 0; F GH = 424.26 lb F GH sin 45°(2) - 150(4) = 0 + © M B = 0: 2 ft 2 ft 3 ft 1 ft 1 ft 1 ft E D C B A I 30  G H Ans: , , M D = - 150 lb # ft V D = - 150 lb N D = 300 lb

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    16 *1–16. A 150-lb bucket is suspended from a cable on the wooden frame. Determine the resultant internal loadings acting on the cross section at E . Support Reactions: We will only need to compute A x , A y , and F BI . Referring to the free-body diagram of the frame, Fig. a , a Internal Loadings: Using the results of A x and A y , section AE of member AB will be considered. Referring to the free-body diagram of this part shown in Fig. b , Ans. Ans. a Ans. The negative sign indicates that N E acts in the opposite sense to that shown on the free-body diagram. M E = 300 lb # ft 100(3) - M E = 0 + © M D = 0; V E = 100 lb 100 - V E = 0 + c © F y = 0; N E = - 323 lb N E + 323.21 = 0 : + © F x = 0; A y = 323.21 lb A y - 200 cos 30° - 150 = 0 + c © F y = 0; A x = 100 lb A x - 200 sin 30° = 0 : + © F x = 0; F BI = 200 lb F BI sin 30°(6) - 150(4) = 0 + © M A = 0; 2 ft 2 ft 3 ft 1 ft 1 ft 1 ft E D C B A I 30  G H

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    17 45  1.5 m 1.5 m 3 m 45  A C B b a a b 5 kN Referring to the FBD of the entire beam, Fig. a , a Referring to the FBD of this segment (section a–a ), Fig. b , Ans. Ans. a Ans. Referring to the FBD (section b–b ) in Fig. c , Ans. Ans. a Ans. M b - b = 3.75 kN # m + © M C = 0; 5.303 sin 45° (3) - 5(1.5) - M b - b = 0 + c © F y = 0; V b - b - 5 sin 45° = 0 V b - b = 3.536 kN = 3.54 kN = - 1.77 kN ; + © F x = 0; N b - b - 5 cos 45° + 5.303 = 0 N b - b = - 1.768 kN + © M C = 0; 5.303 sin 45°(3) - 5(1.5) - M a - a = 0 M a - a = 3.75 kN # m +a © F y ¿ = 0; V a - a + 5.303 sin 45° - 5 = 0 V a - a = 1.25 kN +b© F x ¿ = 0; N a - a + 5.303 cos 45° = 0 N a - a = - 3.75 kN + © M A = 0; N B sin 45°(6) - 5(4.5) = 0 N B = 5.303 kN 1–17. Determine resultant internal loadings acting on section a–a and section b–b . Each section passes through the centerline at point C . Ans: M b - b = 3.75 kN # m V b - b = 3.54 kN # m, N b - b = - 1.77 kN, M a - a = 3.75 kN # m, V a - a = 1.25 kN, N a - a = - 3.75 kN,

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    18 Segment AC: Ans. Ans. a Ans. + © M C = 0; M C + 80(6) = 0; M C = - 480 lb # in. + c © F y = 0; V C = 0 : + © F x = 0; N C + 80 = 0; N C = - 80 lb 1–18. The bolt shank is subjected to a tension of 80 lb. Determine the resultant internal loadings acting on the cross section at point C . A B C 90  6 in. Ans: M C = - 480 lb # in. V C = 0, N C = - 80 lb,

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    19 Referring to the FBD of the entire beam, Fig. a , a Referring to the FBD of this segment, Fig. b , Ans. Ans. a Ans. M C = 31.5 kip # ft + © M C = 0; M C + (3)(3)(1.5) + 1 2 (3)(3)(2) - 18.0(3) = 0 + c © F y = 0; 18.0 - 1 2 (3)(3) - (3)(3) - V C = 0 V C = 4.50 kip : + © F x = 0; N C = 0 + © M B = 0; 1 2 (6)(6)(2) + 1 2 (6)(6)(10) - A y (12) = 0 A y = 18.0 kip 1–19. Determine the resultant internal loadings acting on the cross section through point C . Assume the reactions at the supports A and B are vertical. 3 ft 3 ft D C A B 6 ft 6 kip / ft 6 kip / ft Ans: , , M C = 31.5 kip # ft V C = 4.50 kip N C = 0

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    20 Referring to the FBD of the entire beam, Fig. a , a Referring to the FBD of this segment, Fig. b , Ans. Ans. a Ans. + © M A = 0; M D - 18.0 (2) = 0 M D = 36.0 kip # ft + c © F y = 0; 18.0 - 1 2 (6)(6) - V D = 0 V D = 0 : + © F x = 0; N D = 0 + © M B = 0; 1 2 (6)(6)(2) + 1 2 (6)(6)(10) - A y (12) = 0 A y = 18.0 kip *1–20. Determine the resultant internal loadings acting on the cross section through point D . Assume the reactions at the supports A and B are vertical. 3 ft 3 ft D C A B 6 ft 6 kip / ft 6 kip / ft

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    21 Internal Loadings: Referring to the free-body diagram of the section of the clamp shown in Fig. a , Ans. Ans. a Ans. + © M A = 0; 900(0.2) - M a - a = 0 M a - a = 180 N # m © F x ¿ = 0; V a - a - 900 sin 30° = 0 V a - a = 450 N © F y ¿ = 0; 900 cos 30° - N a - a = 0 N a - a = 779 N 1–21. The forged steel clamp exerts a force of N on the wooden block. Determine the resultant internal loadings acting on section a–a passing through point A . F = 900 200 mm a a F  900 N F  900 N 30  A Ans: M a - a = 180 N # m: V a - a = 450 N, N a - a = 779 N,

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    22 1–22. The metal stud punch is subjected to a force of 120 N on the handle. Determine the magnitude of the reactive force at the pin A and in the short link BC . Also, determine the internal resultant loadings acting on the cross section passing through the handle arm at D . Member: a Ans. Ans. Segment: Ans. Ans. a Ans. M D = 36.0 N # m M D - 120(0.3) = 0 + © M D = 0; V D = 0 + Q© F y ¿ = 0; N D = 120 N N D - 120 = 0 a + © F x ¿ = 0; = 1.49 kN = 1491 N F A = 2 1489.56 2 + 60 2 A x = 60 N A x - 120 sin 30° = 0; ; + © F x = 0; A y = 1489.56 N A y - 1385.6 - 120 cos 30° = 0 + c © F y = 0; F BC = 1385.6 N = 1.39 kN F BC cos 30°(50) - 120(500) = 0 +© M A = 0; 60  50 mm 100 mm 200 mm 300 mm B C D 120 N 50 mm 100 mm E 30  A Ans: , , M D = 36.0 N # m V D = 0, N D = 120 N, F A = 1.49 kN F BC = 1.39 kN

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    23 1–23. Solve Prob. 1–22 for the resultant internal loadings acting on the cross section passing through the handle arm at E and at a cross section of the short link BC . Member: a Segment: Ans. Ans. a Ans. Short link: Ans. Ans. a Ans. M = 0 + © M H = 0; N = 1.39 kN 1.3856 - N = 0; + c © F y = 0; V = 0 ; + © F x = 0; M E = 48.0 N # m M E - 120(0.4) = 0; + © M E = 0; V E = 120 N V E - 120 = 0; a+ © F y ¿ = 0; N E = 0 + b© F x ¿ = 0; F BC = 1385.6 N = 1.3856 kN F BC cos 30°(50) - 120(500) = 0 + © M A = 0; 60  50 mm 100 mm 200 mm 300 mm B C D 120 N 50 mm 100 mm E 30  A Ans: , , Short link: M = 0 N = 1.39 kN, V = 0, M E = 48.0 N # m, V E = 120 N N E = 0

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    24 *1–24. Determine the resultant internal loadings acting on the cross section of the semicircular arch at C . a B y Ans. Ans. a Ans. M C = 0 w 0 r ( r ) - M C + ( - w 0 r )( r ) = 0 + © M 0 = 0; V 2 = 0 w 0 r + V C - w 0 r ( - cos u ) L p 2 0 = 0; w 0 r + V C - w 0 r L p 2 0 sin u d u = 0 + c © F y = 0; N C = - w 0 r sin u L p 2 0 = - w 0 r - N C - w 0 r L p 2 0 cos u d u = 0 : + © F x = 0; = w 0 r B y (2 r ) - w 0 r 2 ( - cos u )] 0 p = 0 B y (2 r ) - w 0 r 2 L p 0 sin u d u = 0 - L p 0 ( w 0 r d u )(sin u ) r (1 - cos u ) = 0 + © M A = 0; B y (2 r ) - L p 0 ( w 0 r d u )(cos u )( r sin u ) C A B w 0 u r

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    25 Ans. Ans. Ans. Ans. Ans. Ans. © M z = 0; ( T B ) z - 105(0.5) = 0; ( T B ) z = 52.5 lb # ft © M y = 0; ( M B ) y - 105(7.5) = 0; ( M B ) y = 788 lb # ft © M x = 0; ( M B ) x = 0 © F z = 0; ( N B ) z = 0 © F y = 0; ( V B ) y = 0 © F x = 0; ( V B ) x - 105 = 0; ( V B ) x = 105 lb 1–25. Determine the resultant internal loadings acting on the cross section through point B of the signpost. The post is fixed to the ground and a uniform pressure of 7 > acts perpendicular to the face of the sign. ft 2 lb 4 ft z y 6 ft x B A 3 ft 2 ft 3 ft 7 lb / ft 2 Ans: , , , , ( T B ) z = 52.5 lb # ft ( M B ) y = 788 lb # ft ( M B ) x = 0 ( N B ) z = 0 ( V B ) y = 0 ( V B ) x = 105 lb,

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    26 1–26. The shaft is supported at its ends by two bearings A and B and is subjected to the forces applied to the pulleys fixed to the shaft. Determine the resultant internal loadings acting on the cross section located at point C . The 300-N forces act in the z direction and the 500-N forces act in the x direction. The journal bearings at A and B exert only x and z components of force on the shaft. + - y B C 400 mm 150 mm 200 mm 250 mm A x z 300 N 300 N 500 N 500 N Ans. Ans. Ans. Ans. Ans. Ans. © M z = 0; ( M C ) z - 1000(0.2) + 750(0.45) = 0; ( M C ) z = - 138 N # m © M y = 0; ( T C ) y = 0 © M x = 0; ( M C ) x + 240(0.45) = 0; ( M C ) x = - 108 N # m © F z = 0; ( V C ) z + 240 = 0; ( V C ) z = - 240 N © F y = 0; ( N C ) y = 0 © F x = 0; ( V C ) x + 1000 - 750 = 0; ( V C ) x = - 250 N Ans: , , , , ( M C ) z = - 138 N # m ( T C ) y = 0 ( M C ) x = - 108 N # m, ( V C ) z = - 240 N ( N C ) y = 0 ( V C ) x = - 250 N

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    27 1–27. The pipe assembly is subjected to a force of 600 N at B . Determine the resultant internal loadings acting on the cross section at C . Internal Loading: Referring to the free-body diagram of the section of the pipe shown in Fig. a , Ans. Ans. Ans. Ans. Ans. Ans. The negative signs indicate that and act in the opposite sense to that shown on the free-body diagram. ( M C ) z ( V C ) y , ( V C ) z , ( T C ) x , ( M C ) z = - 99.0 N # m ( M C ) z + 600 cos 60° cos 30°(0.15) + 600 cos 60° sin 30°(0.4) = 0 © M z = 0; ( M C ) y = 153 N # m ( M C ) y - 600 sin 60° (0.15) - 600 cos 60° sin 30°(0.5) = 0 © M y = 0; ( T C ) x = - 77.9 N # m ( T C ) x + 600 sin 60°(0.4) - 600 cos 60° cos 30°(0.5) = 0 © M x = 0; ( V C ) z = - 520 N ( V C ) z + 600 sin 60° = 0 © F z = 0; ( V C ) y = - 260 N ( V C ) y + 600 cos 60° cos 30° = 0 © F y = 0; ( N C ) x = 150 N ( N C ) x - 600 cos 60° sin 30° = 0 © F x = 0; A C B y x z 400 mm 150 mm 500 mm 600 N 150 mm 30  60  Ans: Ans: , , , , , , ( M C ) z = - 99.0 N # m ( M C ) y = 153 N # m, ( T C ) x = - 77.9 N # m, ( V C ) z = - 520 N ( V C ) y = - 260 N ( N C ) x = 150 N

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    28 Internal Loading: Referring to the free-body diagram of the section of the drill and brace shown in Fig. a , Ans. Ans. Ans. Ans. Ans. Ans. The negative sign indicates that (M A ) z acts in the opposite sense to that shown on the free-body diagram. © M z = 0; A M A B z + 30(1.25) = 0 A M A B z = - 37.5 lb # ft © M y = 0; A T A B y - 30(0.75) = 0 A T A B y = 22.5 lb # ft © M x = 0; A M A B x - 10(2.25) = 0 A M A B x = 22.5 lb # ft © F z = 0; A V A B z - 10 = 0 A V A B z = 10 lb © F y = 0; A N A B y - 50 = 0 A N A B y = 50 lb © F x = 0; A V A B x - 30 = 0 A V A B x = 30 lb *1–28. The brace and drill bit is used to drill a hole at O . If the drill bit jams when the brace is subjected to the forces shown, determine the resultant internal loadings acting on the cross section of the drill bit at A . z x y A O 9 in. 6 in. 6 in. 6 in. 9 in. 3 in. F x  30 lb F y  50 lb F z  10 lb

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    29 1–29. The curved rod AD of radius r has a weight per length of w . If it lies in the vertical plane, determine the resultant internal loadings acting on the cross section through point B . Hint : The distance from the centroid C of segment AB to point O is OC = [2 r sin ( u > 2)] > u . Ans. Ans. a Ans. M B = w r 2 ( u cos u - sin u ) M B = - N B r - wr 2 2 sin ( u /2) cos ( u /2) w r u a cos u 2 b a 2 r sin ( u /2) u b + ( N B ) r + M B = 0 + © M 0 = 0; V B = - w r u sin u - V B - w r u sin u = 0 + Q© F y = 0; N B = - w r u cos u N B + w r u cos u = 0 R + © F x = 0; O r C B D A u u 2 Ans: Ans: , , M B = w r 2 ( u cos u - sin u ) V B = - w r u sin u N B = - w r u cos u

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    30 (1) (2) (3) (4) Since is can add, then , Eq. (1) becomes Neglecting the second order term, QED Eq. (2) becomes Neglecting the second order term, QED Eq. (3) becomes Neglecting the second order term, QED Eq. (4) becomes Neglecting the second order term, QED dM d u = - T Td u + dM = 0 Td u + dM + dTd u 2 = 0 dT d u = M Md u - dT = 0 Md u - dT + dMd u 2 = 0 dV d u = - N Nd u + dV = 0 Nd u + dV + dNd u 2 = 0 dN d u = V Vd u - dN = 0 Vd u - dN + dVd u 2 = 0 cos d u 2 = 1 sin d u 2 = d u 2 d u 2 T sin d u 2 - M cos d u 2 + ( T + dT ) sin d u 2 + ( M + dM ) cos d u 2 = 0 © M y = 0; T cos d u 2 + M sin d u 2 - ( T + dT ) cos d u 2 + ( M + dM ) sin d u 2 = 0 © M x = 0; N sin d u 2 - V cos d u 2 + ( N + dN ) sin d u 2 + ( V + dV ) cos d u 2 = 0 © F y = 0; N cos d u 2 + V sin d u 2 - ( N + dN ) cos d u 2 + ( V + dV ) sin d u 2 = 0 © F x = 0; 1–30. A differential element taken from a curved bar is shown in the figure. Show that and dT > d u = M . dM > d u = - T , dV > d u = - N , dN > d u = V , M V N d u M  dM T  dT N  dN V  dV T

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