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Mechanics of Materials, 10th Edition Solution Manual

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    Solution 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 . A E D B C 4 ft 400 lb 800 lb 4 ft 4 ft 4 ft Solution 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 + Σ M B = 0; C y (8) + 400(4) - 800(12) = 0 C y = 1000 lb Internal Loadings: Using the result for C y , section DE of the shaft will be considered. Referring to the free-body diagram, Fig. b , S + Σ F x = 0; N E = 0 Ans. + c Σ F y = 0; V E + 1000 - 800 = 0 V E = - 200 lb Ans. a + Σ M E = 0; 1000(4) - 800(8) - M E = 0 M E = - 2400 lb # ft = - 2.40 kip # ft Ans. The negative signs indicates that V E and M E act in the opposite sense to that shown on the free-body diagram. Ans: N E = 0, V E = - 200 lb, M E = - 2.40 kip # ft

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    Solution 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 the centroid A . The 500-lb load is applied along the centroidal axis of the member. 30 ° A b a b a 500 lb 500 lb Solution (a) + S Σ F x = 0; N a - 500 = 0 N a = 500 lb Ans. + T Σ F y = 0; V a = 0 Ans. (b) R + Σ F x = 0; N b - 500 cos 30 ° = 0 N b = 433 lb Ans. + Q Σ F y = 0; V b - 500 sin 30 ° = 0 V b = 250 lb Ans. Ans: (a) N a = 500 lb, V a = 0, (b) N b = 433 lb, V b = 250 lb

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    Solution 3 1–3. Determine the resultant internal loadings acting on section b – b through the centroid C on the beam. 30 ° 6 ft 3 ft 60 ° b b 900 lb / ft C A B Solution Support Reaction: a + Σ M A = 0; N B (9 sin 30 ° ) - 1 2 (900)(9)(3) = 0 N B = 2700 lb Equations of Equilibrium: For section b–b S + Σ F x = 0; V b - b + 1 2 (300)(3) sin 30 ° - 2700 = 0 V b - b = 2475 lb = 2.475 kip Ans. + c Σ F y = 0; N b - b - 1 2 (300)(3) cos 30 ° = 0 N b - b = 389.7 lb = 0.390 kip Ans. a + Σ M C = 0; 2700(3 sin 30 ° ) - 1 2 (300)(3)(1) - M b - b = 0 M b - b = 3600 lb # ft = 3.60 kip # ft Ans. Ans: V b - b = 2.475 kip, N b - b = 0.390 kip, M b - b = 3.60 kip # ft

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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 . A D B C 900 N 1.5 m 600 N / m 1.5 m 1 m 1 m 1 m Solution 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 + Σ M A = 0; B y (4.5) - 600(2)(2) - 900(6) = 0 B y = 1733.33 N 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 , S + Σ F x = 0; N C = 0 Ans. + c Σ F y = 0; V C - 600(1) + 1733.33 - 900 = 0 V C = - 233 N Ans. a + Σ M C = 0; 1733.33(2.5) - 600(1)(0.5) - 900(4) - M C = 0 M C = 433 N # m Ans. The negative sign indicates that V C acts in the opposite sense to that shown on the free-body diagram. * Ans: N C = 0, V C = - 233 N, M C = 433 N # m

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    5 1–5. Determine the resultant internal loadings acting on the cross section at point B . A C 12 ft 3 ft 60 lb / ft B Solution S + Σ F x = 0; N B = 0 Ans. + c Σ F y = 0; V B - 1 2 (48)(12) = 0 V B = 288 lb Ans. a + Σ M B = 0; - M B - 1 2 (48)(12)(4) = 0 M B = - 1152 lb # ft = - 1.15 kip # ft Ans. Ans: N B = 0 V B = 288 lb, M B = - 1.15 kip # ft ,

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    6 1–6. Determine the resultant internal loadings on the cross section at point D . 2 m D 1.25 kN / m 1.5 m 1 m E B C A F 0.5 m 0.5 m 0.5 m Solution Support Reactions: Member BC is the two force member. a + Σ M A = 0; 4 5 F BC (1.5) - 1.875(0.75) = 0 F BC = 1.1719 kN + c Σ F y = 0; A y + 4 5 (1.1719) - 1.875 = 0 A y = 0.9375 kN S + Σ F x = 0; 3 5 (1.1719) - A x = 0 A x = 0.7031 kN Equations of Equilibrium: For point D S + Σ F x = 0; N D - 0.7031 = 0 N D = 0.703 kN Ans. + c Σ F y = 0; 0.9375 - 0.625 - V D = 0 V D = 0.3125 kN Ans. a + Σ M D = 0; M D + 0.625(0.25) - 0.9375(0.5) = 0 M D = 0.3125 kN # m Ans. Ans: N D = 0.703 kN, V D = 0.3125 kN, M D = 0.3125 kN # m

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    7 1–7. Determine the resultant internal loadings at cross sections at points E and F on the assembly. 2 m D 0.5 m 0.5 m 1.25 kN / m 0.5 m 1.5 m 1 m E B C A F Solution Support Reactions: Member BC is the two-force member. a + Σ M A = 0; 4 5 F BC (1.5) - 1.875(0.75) = 0 F BC = 1.1719 kN + c Σ F y = 0; A y + 4 5 (1.1719) - 1.875 = 0 A y = 0.9375 kN S + Σ F x = 0; 3 5 (1.1719) - A x = 0 A x = 0.7031 kN Equations of Equilibrium: For point F + b Σ F x ′ = 0; N F - 1.1719 = 0 N F = 1.17 kN Ans. a + Σ F y ′ = 0; V F = 0 Ans. a + Σ M F = 0; M F = 0 Ans. Equations of Equilibrium: For point E + d Σ F x = 0; N E - 3 5 (1.1719) = 0 N E = 0.703 kN Ans. + c Σ F y = 0; V E - 0.625 + 4 5 (1.1719) = 0 V E = - 0.3125 kN Ans. a + Σ M E = 0; - M E - 0.625(0.25) + 4 5 (1.1719)(0.5) = 0 M E = 0.3125 kN # m Ans. Negative sign indicates that V E acts in the opposite direction to that shown on FBD. Ans: N F = 1.17 kN, V F = 0, M F = 0, N E = 0.703 kN, V E = - 0.3125 kN, M E = 0.3125 kN # m

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    8 * 1–8. The beam supports the distributed load shown. Determine the resultant internal loadings acting on the cross section at point C . Assume the reactions at the supports A and B are vertical. 1.5 m 3 m D C A B 4 kN / m 1.5 m Solution Support Reactions: Referring to the FBD of the entire beam, Fig. a , a + Σ M A = 0; B y (6) - 1 2 (4)(6)(2) = 0 B y = 4.00 kN Internal Loadings: Referring to the FBD of the right segment of the beam sectioned through C , Fig. b , S + Σ F x = 0; N C = 0 Ans. + c Σ F y = 0; V C + 4.00 - 1 2 (3)(4.5) = 0 V C = 2.75 kN Ans. a + Σ M C = 0; 4.00(4.5) - 1 2 (3)(4.5)(1.5) - M C = 0 M C = 7.875 kN # m Ans. Ans: N C = 0, V C = 2.75 kN, M C = 7.875 kN # m

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    9 1–9. The beam supports the distributed load shown. Determine the resultant internal loadings acting on the cross section at point D . Assume the reactions at the supports A and B are vertical. 1.5 m 3 m D C A B 4 kN / m 1.5 m Solution Support Reactions: Referring to the FBD of the entire beam, Fig. a , a + Σ M A = 0; B y (6) - 1 2 (4)(6)(2) = 0 B y = 4.00 kN Internal Loadings: Referring to the FBD of the right segment of the beam sectioned through D , Fig. b , S + Σ F x = 0; N D = 0 Ans. + c Σ F y = 0; V D + 4.00 - 1 2 (1.00)(1.5) = 0 V D = - 3.25 kN Ans. a + Σ M D = 0; 4.00(1.5) - 1 2 (1.00)(1.5)(0.5) - M D = 0 M D = 5.625 kN # m Ans. The negative sign indicates that V D acts in the sense opposite to that shown on the FBD. Ans: N D = 0, V D = - 3.25 kN, M D = 5.625 kN # m

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

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    11 1–11. Determine the resultant internal loadings acting on the cross sections at points D and E of the frame. 4 ft A D 2 ft C B 75 lb / ft 1 ft 150 lb 1 ft E 30 ° 2 ft 1 ft G 1 ft F Solution Member AG : a + Σ M A = 0; 4 5 F BC (3) - 75(4)(5) - 150 cos 30 ° (7) = 0; F BC = 1003.89 lb a + Σ M B = 0; A y (3) - 75(4)(2) - 150 cos 30 ° (4) = 0; A y = 373.20 lb S + Σ F x = 0; A x - 3 5 (1003.89) + 150 sin 30 ° = 0; A x = 527.33 lb For point D : S + Σ F x = 0; N D + 527.33 = 0 N D = - 527 lb Ans. + c Σ F y = 0; - 373.20 - V D = 0 V D = - 373 lb Ans. a + Σ M D = 0; M D + 373.20(1) = 0 M D = - 373 lb # ft Ans. For point E : S + Σ F x = 0; 150 sin 30 ° - N E = 0 N E = 75.0 lb Ans. + c Σ F y = 0; V E - 75(3) - 150 cos 30 ° = 0 V E = 355 lb Ans. a + Σ M E = 0; - M E - 75(3)(1.5) - 150 cos 30 ° (3) = 0; M E = - 727 lb # ft Ans. Ans: N D = - 527 lb, V D = - 373 lb, M D = - 373 lb ft, # N E = 75.0 lb, V E = 355 lb , M E = - 727 lb # ft

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    12 1–12. Determine the resultant internal loadings acting on the cross sections at points F and G of the frame. 4 ft A D 2 ft C B 75 lb / ft 1 ft 150 lb 1 ft E 30 ° 2 ft 1 ft G 1 ft F Solution Member AG : a + Σ M A = 0; 4 5 F BF (3) - 300(5) - 150 cos 30 ° (7) = 0 F BF = 1003.9 lb For point F : + Q Σ F x ′ = 0; V F = 0 Ans. a + Σ F y ′ = 0; N F - 1003.9 = 0 N F = 1004 lb Ans. a + Σ M F = 0; M F = 0 Ans. For point G : d + Σ F x = 0; N G - 150 sin 30 ° = 0 N G = 75.0 lb Ans. + c Σ F y = 0; V G - 75(1) - 150 cos 30 ° = 0 V G = 205 lb Ans. a + Σ M G = 0; - M G - 75(1)(0.5) - 150 cos 30 ° (1) = 0 M G = - 167 lb # ft Ans. * Ans: V F = 0, N F = 1004 lb, M F = 0 , N G = 75.0 lb, V G = 205 lb, M G = - 167 lb # ft

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

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

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    15 1–15. The beam supports the triangular distributed load shown. Determine the resultant internal loadings on the cross section at point C . Assume the reactions at the supports A and B are vertical. 6 ft 6 ft C A B 4.5 ft 800 lb / ft 6 ft 4.5 ft E D Solution Support Reactions: Referring to the FBD of the entire beam, Fig. a , a + Σ M B = 0; 1 2 (0.8)(18)(6) - 1 2 (0.8)(9)(3) - A y (18) = 0 A y = 1.80 kip Internal Loadings: Referring to the FBD of the left beam segment sectioned through point C , Fig. b , S + Σ F x = 0; N C = 0 Ans. + c Σ F y = 0; 1.80 - 1 2 (0.5333)(12) - V C = 0 V C = - 1.40 kip Ans. a + Σ M C = 0; M C + 1 2 (0.5333)(12)(4) - 1.80(12) = 0 M C = 8.80 kip # ft Ans. The negative sign indicates that V C acts in the sense opposite to that shown on the FBD. Ans: N C = 0, V C = - 1.40 kip, M C = 8.80 kip # ft

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