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https://gioumeh.com/product/mechanics-of-materials-solution/ click here to download @solutionmanual1 © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. https://gioumeh.com/product/mechanics-of-materials-solution/ 1–1. The shaft is supported by a smooth thrust bearing and clickat Bhere a journal bearing at C. Determine the resultant internal loadings acting on the cross section at E. to download B A 4 ft C E 4 ft 4 ft D 4 ft 400 lb 800 lb Solution Support Reactions: We will only need to compute Cy by writing the moment equation of equilibrium about B with reference to the free-body diagram of the entire shaft, Fig. a. a+ ΣMB = 0; Cy(8) + 400(4) - 800(12) = 0 Cy = 1000 lb Internal Loadings: Using the result for Cy, section DE of the shaft will be considered. Referring to the free-body diagram, Fig. b, + ΣFx = 0; NE = 0 S Ans. + c ΣFy = 0; VE + 1000 - 800 = 0 VE = -200 lb Ans. a+ ΣME = 0; 1000(4) - 800(8) - ME = 0 ME = - 2400 lb # ft = - 2.40 kip # ft Ans. The negative signs indicates that VE and ME act in the opposite sense to that shown on the free-body diagram. @solutionmanual1 1 Ans: NE = 0, VE = -200 lb, ME = - 2.40 kip # ft © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. https://gioumeh.com/product/mechanics-of-materials-solution/ 1–2. Determine the resultant internal normal and shear forcehere in click 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. to download a b 30 500 lb 500 lb b A a Solution (a) + ΣFx = 0; S Na - 500 = 0 Ans. Na = 500 lb + T Σ Fy = 0; Va = 0 Ans. (b) R+ ΣFx = 0; Nb - 500 cos 30° = 0 Ans. Nb = 433 lb +Q ΣFy = 0; Vb - 500 sin 30° = 0 Ans. Vb = 250 lb Ans: (a) Na = 500 lb, Va = 0, (b) Nb = 433 lb, Vb = 250 lb @solutionmanual1 2 © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. https://gioumeh.com/product/mechanics-of-materials-solution/ 1–3. Determine the resultant internal loadings acting on section click here b–b through the centroid C on the beam. to download B b 900 lb/ft b C A 30 60 3 ft 6 ft Solution Support Reaction: a+ ΣMA = 0; NB(9 sin 30°) - 1 (900)(9)(3) = 0 2 NB = 2700 lb Equations of Equilibrium: For section b–b 1 + ΣFx = 0; Vb - b + (300)(3) sin 30° - 2700 = 0 S 2 Vb - b = 2475 lb = 2.475 kip Ans. 1 (300)(3) cos 30° = 0 2 Nb - b = 389.7 lb = 0.390 kip Ans. + c ΣFy = 0; Nb - b - a+ ΣMC = 0; 2700(3 sin 30°) 1 - (300)(3)(1) - Mb - b = 0 2 Mb - b = 3600 lb # ft = 3.60 kip # ft Ans. Ans: Vb - b = 2.475 kip, Nb - b = 0.390 kip, Mb - b = 3.60 kip # ft @solutionmanual1 3 © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. https://gioumeh.com/product/mechanics-of-materials-solution/ *1–4. The shaft is supported by a smooth thrust bearing at A and click here a smooth journal bearing at B. Determine the resultant internal loadings acting on the cross section at C. to download 600 N/m A B D C 1m 1m 1m 1.5 m 1.5 m 900 N Solution Support Reactions: We will only need to compute By by writing the moment equation of equilibrium about A with reference to the free-body diagram of the entire shaft, Fig. a. a+ ΣMA = 0; By(4.5) - 600(2)(2) - 900(6) = 0 By = 1733.33 N Internal Loadings: Using the result of By, section CD of the shaft will be considered. Referring to the free-body diagram of this part, Fig. b, + ΣFx = 0; S Ans. NC = 0 + c ΣFy = 0; VC - 600(1) + 1733.33 - 900 = 0 a+ ΣMC = 0; 1733.33(2.5) - 600(1)(0.5) - 900(4) - MC = 0 VC = -233 N MC = 433 N # m Ans. Ans. The negative sign indicates that VC acts in the opposite sense to that shown on the free-body diagram. Ans: NC = 0, VC = - 233 N, MC = 433 N # m @solutionmanual1 4 © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. https://gioumeh.com/product/mechanics-of-materials-solution/ 1–5. Determine the resultant internal loadings acting the clickonhere cross section at point B. to download60 lb/ ft A C B 3 ft 12 ft Solution + ΣFx = 0; S + c ΣFy = 0; Ans. NB = 0 VB - 1 (48)(12) = 0 2 Ans. VB = 288 lb a+ ΣMB = 0; -MB - 1 (48)(12)(4) = 0 2 MB = -1152 lb # ft = - 1.15 kip # ft Ans. @solutionmanual1 5 Ans: NB = 0, VB = 288 lb, MB = - 1.15 kip # ft © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. https://gioumeh.com/product/mechanics-of-materials-solution/ 1–6. Determine the resultant internal loadings on click the cross here section at point D. to download C 1m F 2m 1.25 kN/m Solution A Support Reactions: Member BC is the two force member. 4 a+ ΣMA = 0; FBC (1.5) - 1.875(0.75) = 0 5 D E 0.5 m 0.5 m 0.5 m B 1.5 m FBC = 1.1719 kN + c ΣFy = 0; Ay + 4 (1.1719) - 1.875 = 0 5 Ay = 0.9375 kN 3 + ΣFx = 0; (1.1719) - Ax = 0 S 5 Ax = 0.7031 kN Equations of Equilibrium: For point D + ΣFx = 0; ND - 0.7031 = 0 S ND = 0.703 kN Ans. + c ΣFy = 0; 0.9375 - 0.625 - VD = 0 VD = 0.3125 kN Ans. a+ ΣMD = 0; MD + 0.625(0.25) - 0.9375(0.5) = 0 MD = 0.3125 kN # m Ans. Ans: ND = 0.703 kN, VD = 0.3125 kN, MD = 0.3125 kN # m @solutionmanual1 6 © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. https://gioumeh.com/product/mechanics-of-materials-solution/ 1–7. Determine the resultant internal loadings at click cross sections here at points E and F on the assembly. to download C 1m F 2m 1.25 kN/m Solution A Support Reactions: Member BC is the two-force member. 4 a+ ΣMA = 0; FBC (1.5) - 1.875(0.75) = 0 5 D E 0.5 m 0.5 m 0.5 m B 1.5 m FBC = 1.1719 kN + c ΣFy = 0; Ay + 4 (1.1719) - 1.875 = 0 5 Ay = 0.9375 kN 3 + ΣFx = 0; (1.1719) - Ax = 0 S 5 Ax = 0.7031 kN Equations of Equilibrium: For point F + bΣFx′ = 0; NF - 1.1719 = 0 NF = 1.17 kN Ans. a + ΣFy′ = 0; VF = 0 Ans. a+ ΣMF = 0; MF = 0 Ans. Equations of Equilibrium: For point E + ΣFx = 0; NE - 3 (1.1719) = 0 d 5 Ans. NE = 0.703 kN + c ΣFy = 0; VE - 0.625 + 4 (1.1719) = 0 5 Ans. VE = - 0.3125 kN a+ ΣME = 0; - ME - 0.625(0.25) + 4 (1.1719)(0.5) = 0 5 ME = 0.3125 kN # m Ans. Negative sign indicates that VE acts in the opposite direction to that shown on FBD. Ans: NF = 1.17 kN, VF = 0, MF = 0, NE = 0.703 kN, VE = - 0.3125 kN, ME = 0.3125 kN # m @solutionmanual1 7 © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. https://gioumeh.com/product/mechanics-of-materials-solution/ *1–8. The beam supports the distributed load shown. Determine click here the resultant internal loadings acting on the cross section at point C. Assume the reactions at the supports A and B are vertical. 4 kN/m to download A B D C 1.5 m 3m 1.5 m Solution Support Reactions: Referring to the FBD of the entire beam, Fig. a, a+ ΣMA = 0; By(6) - 1 (4)(6)(2) = 0 2 By = 4.00 kN Internal Loadings: Referring to the FBD of the right segment of the beam sectioned through C, Fig. b, + ΣFx = 0; NC = 0 S + c ΣFy = 0; VC + 4.00 a+ ΣMC = 0; 4.00(4.5) - Ans. 1 (3)(4.5) = 0 2 Ans. VC = 2.75 kN 1 (3)(4.5)(1.5) - MC = 0 2 MC = 7.875 kN # m Ans. Ans: NC = 0, VC = 2.75 kN, MC = 7.875 kN # m @solutionmanual1 8 © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. https://gioumeh.com/product/mechanics-of-materials-solution/ 1–9. The beam supports the distributed load shown. Determine click here the resultant internal loadings acting on the cross section at point D. Assume the reactions at the supports A and B are vertical. 4 kN/m to download A B D C 1.5 m 3m 1.5 m Solution Support Reactions: Referring to the FBD of the entire beam, Fig. a, a+ ΣMA = 0; By(6) - 1 (4)(6)(2) = 0 2 By = 4.00 kN Internal Loadings: Referring to the FBD of the right segment of the beam sectioned through D, Fig. b, + ΣFx = 0; ND = 0 S Ans. + c ΣFy = 0; VD + 4.00 a+ ΣMD = 0; 4.00(1.5) - 1 (1.00)(1.5) = 0 2 VD = -3.25 kN Ans. 1 (1.00)(1.5)(0.5) - MD = 0 2 MD = 5.625 kN # m Ans. The negative sign indicates that VD acts in the sense opposite to that shown on the FBD. Ans: ND = 0, VD = -3.25 kN, MD = 5.625 kN # m @solutionmanual1 9 © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. https://gioumeh.com/product/mechanics-of-materials-solution/ 1–10. The boom DF of the jib crane and the column DE havehere a click 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. to download D 2 ft F A B 8 ft 3 ft 5 ft C 300 lb 7 ft Solution E Equations of Equilibrium: For point A + ΣFx = 0; d + c ΣFy = 0; VA - 150 - 300 = 0 a+ ΣMA = 0; Ans. NA = 0 Ans. VA = 450 lb -MA - 150(1.5) - 300(3) = 0 MA = - 1125 lb # ft = - 1.125 kip # ft Ans. Negative sign indicates that MA acts in the opposite direction to that shown on FBD. Equations of Equilibrium: For point B + ΣFx = 0; d + c ΣFy = 0; VB - 550 - 300 = 0 a+ ΣMB = 0; Ans. NB = 0 Ans. VB = 850 lb - MB - 550(5.5) - 300(11) = 0 MB = -6325 lb # ft = -6.325 kip # ft Ans. Negative sign indicates that MB acts in the opposite direction to that shown on FBD. Equations of Equilibrium: For point C + ΣFx = 0; d + c ΣFy = 0; Ans. VC = 0 - NC - 250 - 650 - 300 = 0 Ans. NC = -1200 lb = - 1.20 kip a+ ΣMC = 0; -MC - 650(6.5) - 300(13) = 0 MC = -8125 lb # ft = -8.125 kip # ft Ans. Negative signs indicate that NC and MC act in the opposite direction to that shown on FBD. @solutionmanual1 10 Ans: NA = 0, VA = 450 lb, MA = -1.125 kip # ft, NB = 0, VB = 850 lb, MB = - 6.325 kip # ft, VC = 0, NC = -1.20 kip, MC = -8.125 kip # ft © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. https://gioumeh.com/product/mechanics-of-materials-solution/ 1–11. Determine the resultant internal loadings acting the clickonhere cross sections at points D and E of the frame. to download C 1 ft F 4 ft 75 lb/ft A 1 ft Solution Member AG: B 1 ft E 2 ft G 1 ft 30 150 lb a+ ΣMA = 0; 4 FBC (3) - 75(4)(5) - 150 cos 30°(7) = 0; FBC = 1003.89 lb 5 a+ ΣMB = 0; Ay (3) - 75(4)(2) - 150 cos 30°(4) = 0; Ay = 373.20 lb + ΣFx = 0; S D 2 ft Ax - 3 (1003.89) + 150 sin 30° = 0; Ax = 527.33 lb 5 For point D: + ΣFx = 0; S ND + 527.33 = 0 ND = -527 lb + c ΣFy = 0; -373.20 - VD = 0 a+ ΣMD = 0; Ans. Ans. VD = -373 lb MD + 373.20(1) = 0 MD = -373 lb # ft Ans. For point E: + ΣFx = 0; S 150 sin 30° - NE = 0 Ans. NE = 75.0 lb + c ΣFy = 0; VE - 75(3) - 150 cos 30° = 0 VE = 355 lb a+ ΣME = 0; - ME - 75(3)(1.5) - 150 cos 30°(3) = 0; ME = - 727 lb # ft Ans. Ans. @solutionmanual1 11 Ans: ND = -527 lb, VD = - 373 lb, MD = - 373 lb # ft, NE = 75.0 lb, VE = 355 lb, ME = -727 lb # ft © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. https://gioumeh.com/product/mechanics-of-materials-solution/ *1–12. Determine the resultant internal loadings acting on the click here cross sections at points F and G of the frame. to download C 1 ft F 4 ft 75 lb/ft A 1 ft Solution Member AG: a+ ΣMA = 0; D 2 ft B 1 ft E 2 ft G 1 ft 30 150 lb 4 FBF (3) - 300(5) - 150 cos 30°(7) = 0 5 FBF = 1003.9 lb For point F: +Q ΣFx′ = 0; a +ΣFy′ = 0; NF - 1003.9 = 0 NF = 1004 lb a+ ΣMF = 0; Ans. VF = 0 Ans. Ans. MF = 0 For point G: + ΣFx = 0; d NG - 150 sin 30° = 0 NG = 75.0 lb Ans. + c ΣFy = 0; VG - 75(1) - 150 cos 30° = 0 VG = 205 lb a+ ΣMG = 0; - MG - 75(1)(0.5) - 150 cos 30°(1) = 0 Ans. MG = - 167 lb # ft Ans. Ans: VF = 0, NF = 1004 lb, MF = 0, NG = 75.0 lb, VG = 205 lb, MG = -167 lb # ft @solutionmanual1 12 © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. https://gioumeh.com/product/mechanics-of-materials-solution/ 1–13. The blade of the hacksaw is subjected to a pretension force click here of F = 100 N. Determine the resultant internal loadings acting on section a–a that passes through point D. to download a 225 mm 30 b B A D b F E Solution a 150 mm F C Internal Loadings: Referring to the free-body diagram of the section of the hacksaw shown in Fig. a, + ΣFx = 0; d Na - a + 100 = 0 + c ΣFy = 0; Va - a = 0 a+ ΣMD = 0; - Ma - a - 100(0.15) = 0 Na - a = -100 N Ans. Ans. Ma - a = -15 N # m Ans. The negative sign indicates that Na - a and Ma - a act in the opposite sense to that shown on the free-body diagram. @solutionmanual1 13 Ans: Na - a = - 100 N, Va - a = 0, Ma - a = - 15 N # m © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. https://gioumeh.com/product/mechanics-of-materials-solution/ 1–14. The blade of the hacksaw is subjected to a pretension clickforce here of F = 100 N. Determine the resultant internal loadings acting on section b–b that passes through point D. to download a 225 mm 30 b B A D b F E Solution a 150 mm F C Internal Loadings: Referring to the free-body diagram of the section of the hacksaw shown in Fig. a, ΣFx′ = 0; Nb - b + 100 cos 30° = 0 Nb - b = - 86.6 N Ans. ΣFy′ = 0; Vb - b = 50 N Vb - b - 100 sin 30° = 0 Ans. a+ ΣMD = 0; Mb - b = - 15 N # m - Mb - b - 100(0.15) = 0 Ans. The negative sign indicates that Nb–b and Mb–b act in the opposite sense to that shown on the free-body diagram. @solutionmanual1 14 Ans: Nb - b = - 86.6 N, Vb - b = 50 N, Mb - b = -15 N # m © 2017 Pearson Education, Inc., Hoboken, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. https://gioumeh.com/product/mechanics-of-materials-solution/ 1–15. The beam supports the triangular distributedclick load shown. here Determine the resultant internal loadings on the cross section at point C. Assume the reactions at the supports A and B are vertical. to download 800 lb/ft A D 6 ft 6 ft B C 6 ft E 4.5 ft 4.5 ft Solution Support Reactions: Referring to the FBD of the entire beam, Fig. a, 1 1 a+ ΣMB = 0; (0.8)(18)(6) - (0.8)(9)(3) - Ay(18) = 0 Ay = 1.80 kip 2 2 Internal Loadings: Referring to the FBD of the left beam segment sectioned through point C, Fig. b, + ΣFx = 0; S NC = 0 + c ΣFy = 0; 1.80 - a+ ΣMC = 0; MC + Ans. 1 (0.5333)(12) - VC = 0 VC = -1.40 kip 2 Ans. 1 (0.5333)(12)(4) - 1.80(12) = 0 2 MC = 8.80 kip # ft Ans. The negative sign indicates that VC acts in the sense opposite to that shown on the FBD. Ans: NC = 0, VC = -1.40 kip, MC = 8.80 kip # ft @solutionmanual1 15