Hibbeler_Resistencia_dos_Materiais_8edicao-compactado

1 (a) Ans. (b) Ans. F A = 34.9 kN + c © F y = 0; F A - 4.5 - 4.5 - 5.89 - 6 - 6 - 8 = 0 F A = 13.8 kip + c © F y = 0; F A - 1.0 - 3 - 3 - 1.8 - 5 = 0 1\u20131. Determine the resultant internal normal force acting on the cross section through point A in each column. In (a), segment BC weighs 180 > ft and segment CD weighs 250 > ft. In (b), the column has a mass of 200 > m. kg lb lb © 2010 Pearson Education, Inc ., Upper Saddle River , NJ . All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means , without permission in writing from the publishe r. 8 kN 3 m 1 m 6 kN 6 kN 4.5 kN 4.5 kN 200 mm 200 mm A (b) 200 mm 200 mm 3 kip 3 kip 5 kip 10 ft 4 ft 4 ft 8 in. 8 in. A C D (a) B 1\u20132. Determine the resultant internal torque acting on the cross sections through points C and D . The support bearings at A and B allow free turning of the shaft. Ans. Ans. © M x = 0; T D = 0 T C = 250 N # m © M x = 0; T C - 250 = 0 A B D C 300 mm 200 mm 150 mm 200 mm 250 mm 150 mm 400 N \ue002 m 150 N \ue002 m 250 N \ue002 m Ans. Ans. T C = 500 lb # ft © M x = 0; T C - 500 = 0 T B = 150 lb # ft © M x = 0; T B + 350 - 500 = 0 1\u20133. Determine the resultant internal torque acting on the cross sections through points B and C . 3 ft 2 ft 2 ft 1 ft B A C 500 lb \ue002 ft 350 lb \ue002 ft 600 lb \ue002 ft 01 Solutions 46060 5/6/10 2:43 PM Page 1 2 © 2010 Pearson Education, Inc., Upper Saddle River, NJ . All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means , without permission in writing from the publishe r. *1\u20134. A force of 80 N is supported by the bracket as shown. Determine the resultant internal loadings acting on the section through point A . Equations of Equilibrium: Ans. Ans. a Ans. or a Ans. Negative sign indicates that M A acts in the opposite direction to that shown on FBD . M A =- 0.555 N # m - 80 cos 15°(0.1 cos 30°) = 0 + © M A = 0; M A + 80 sin 15°(0.3 + 0.1 sin 30°) M A =- 0.555 N # m - 80 sin 45°(0.1 + 0.3 sin 30°) = 0 + © M A = 0; M A + 80 cos 45°(0.3 cos 30°) V A = 20.7 N a + © F y ¿ = 0; V A - 80 sin 15° = 0 N A = 77.3 N + Q© F x ¿ = 0; N A - 80 cos 15° = 0 0.1 m 0.3 m 30 \ue003 80 N A 45 \ue003 01 Solutions 46060 5/6/10 2:43 PM Page 2 3 Support Reactions: F or member AB a Equations of Equilibrium: F or point D Ans. Ans. a Ans. Equations of Equilibrium: F or 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 \u2022 1\u20135. 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. © 2010 Pearson Education, Inc., Upper Saddle River, NJ . All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means , without permission in writing from the publishe r. 6 ft 4 ft A 4 ft B C DE 6 ft 3 kip 1.5 kip / ft 01 Solutions 46060 5/6/10 2:43 PM Page 3 4 Support Reactions: a Equations of Equilibrium: F or 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 1\u20136. Determine the normal force , shear force, and moment at a section through point C . T ake P = 8 kN . © 2010 Pearson Education, Inc., Upper Saddle River, NJ . All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means , without permission in writing from the publishe r. 0.75 m C P A B 0.5 m 0.1 m 0.75 m 0.75 m Support Reactions: a Ans. Equations of Equilibrium: F or 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\u20137. T he 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 01 Solutions 46060 5/6/10 2:43 PM Page 4 5 © 2010 Pearson Education, Inc., Upper Saddle River, NJ . All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means , without permission in writing from the publishe r. Referring to the FBD of the entire beam, F ig. a , a Referring to the FBD of this segment, F ig. 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\u20138. 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 Referring to the FBD of the entire beam, F ig. a , a Referring to the FBD of this segment, F ig. 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 \u2022 1\u20139. Determine the resultant internal loadings on the cross section through point D . 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 01 Solutions 46060 5/6/10 2:43 PM Page 5 6 © 2010 Pearson Education, Inc., Upper Saddle River, NJ . All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means , without permission in writing from the publishe r. Equations of Equilibrium: F or 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: F or 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: F or 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\u201310. T he 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 01 Solutions 46060 5/6/10 2:43 PM Page 6 7 © 2010 Pearson Education, Inc., Upper Saddle River, NJ . All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means , without permission in writing from the publishe r. Equations of Equilibrium: F or section a\u2013a Ans. Ans. a Ans. M A = 14.5 lb # in. +© M A = 0; - M A - 80 sin 15°(0.16) + 80 cos 15°(0.23) = 0 N A = 20.7 lb a + © F y ¿ = 0; N A - 80 sin 15° = 0 V A = 77.3 lb + Q© F x ¿ = 0; V A - 80 cos 15° = 0 1\u201311. T he force acts on the gear tooth. Determine the resultant internal loadings on the root of the tooth, i.e ., at the centroid point A of section a\u2013a . F = 80 lb a 30 \ue003 a F \ue004 80 lb 0.23 in. 45 \ue003 A 0.16 in. Support Reactions: Equations of Equilibrium: F or point D Ans. Ans. Ans. Equations of Equilibrium: F or point E Ans. Ans. Ans. M E = 18.0 kN # m d +© M E = 0; 90.0(0.2) - M E = 0 + c © F y = 0; N E = 0 V E = 90.0 kN : + © F x = 0; 90.0 - V E = 0 M D = 21.6 kN # m d +© M D = 0; M D + 18(0.3) - 90.0(0.3) = 0 N D = 18.0 kN + c © F y = 0; N D - 18 = 0 V D = 90.0 kN : + © F x = 0; V D - 90.0 = 0 : + © F x = 0; N C - 90.0 = 0 N C = 90.0 kN N A = 90.0 kN d +© M C = 0; 18(0.7) - 18.0(0.2) - N A (0.1) = 0 + c © F y = 0; N B - 18 = 0 N B = 18.0 kN *1\u201312. T he sky hook is used to support the cable of a scaffold over the side of a building. If it consists of a smooth rod that contacts the parapet of a wall at points A , B , and C , determine the normal force , shear force , and moment on the cross section at points D and E. 0.2 m 0.2 m 0.2 m 0.2 m 0.2 m 0.3 m 0.3 m 18 kN A DE B C 01 Solutions 46060 5/6/10 2:43 PM Page 7 8 © 2010 Pearson Education, Inc., Upper Saddle River, NJ . All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means , without permission in writing from the publishe r. \u2022 1\u201313. T he 800-lb load is being hoisted at a constant speed using the motor M , which has a weight of 90 lb . Determine the resultant internal loadings acting on the cross section through point B in the beam. T he beam has a weight of 40 lb > ft and is fixed to the wall at A . M 4 ft 3 ft 4 ft CB 1.5 ft A 0.25 ft 4 ft 3 ft D 1\u201314. Determine the resultant internal loadings acting on the cross section through points C and D of the beam in Prob . 1\u201313. M 4 ft 3 ft 4 ft CB 1.5 ft A 0.25 ft 4 ft 3 ft D Ans. Ans. a Ans. M B =- 3.12 kip # ft + © M B = 0; - M B - 0.16(2) - 0.8(4.25) + 0.4(1.5) = 0 V B = 0.960 kip + c © F y = 0; V B - 0.8 - 0.16 = 0 N B =- 0.4 kip : + © F x = 0; - N B - 0.4 = 0 F or point C: Ans. Ans. a Ans. F or point D: Ans. Ans. a Ans. M D =- 15.7 kip # ft +© M D = 0; - M D - 0.09(4) - 0.04(14)(7) - 0.8(14.25) = 0 + c © F y = 0; V D - 0.09 - 0.04(14) - 0.8 = 0; V D = 1.45 kip ; + © F x = 0; N D = 0 M C =- 6.18 kip # ft + © M C = 0; - M C - 0.8(7.25) - 0.04(7)(3.5) + 0.4(1.5) = 0 + c © F y = 0; V C - 0.8 - 0.04 (7) = 0; V C = 1.08 kip ; + © F x = 0; N C + 0.4 = 0; N C =- 0.4kip 01 Solutions 46060 5/6/10 2:43 PM Page 8 © 2010 Pearson Education, Inc., Upper Saddle River, NJ . All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means , without permission in writing from the publishe r. 1\u201315. Determine the resultant internal loading on the cross section through point C of the pliers . T here is a pin at A , and the jaws at B are smooth. 120 mm 40 mm 15 mm 80 mm A C D 30 \ue003 20 N 20 N B 9 *1\u201316. Determine the resultant internal loading on the cross section through point D of the pliers . 120 mm 40 mm 15 mm 80 mm A C D 30 \ue003 20 N 20 N B Ans. Ans. + d Ans. © M C = 0; - M C + 60(0.015) = 0; M C = 0.9 N.m : + © F x = 0; N C = 0 + c © F y = 0; - V C + 60 = 0; V C = 60 N Ans. Ans. + d Ans. © M D = 0; M D - 20(0.08) = 0; M D = 1.60 N.m +b © F x = 0; N D - 20 sin 30° = 0; N D = 10 N R+ © F y = 0; V D - 20 cos 30° = 0; V D = 17.3 N 01 Solutions 46060 5/6/10 2:43 PM Page 9 10 © 2010 Pearson Education, Inc., Upper Saddle River, NJ . All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means , without permission in writing from the publishe r. 45 \ue003 1.5 m 1.5 m 3 m 45 \ue003 A C B b a a b 5 kN Referring to the FBD of the entire beam, F ig. a , a Referring to the FBD of this segment (section a\u2013a ), F ig. b , Ans. Ans. a Ans. Referring to the FBD (section b\u2013b ) in F ig. 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 \u2022 1\u201317. Determine resultant internal loadings acting on section a\u2013a and section b\u2013b . Each section passes through the centerline at point C . 01 Solutions 46060 5/6/10 2:43 PM Page 10

Hibbeler_Resistencia_dos_Materiais_8edicao-compactado

Mecânica dos Sólidos I

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Mecânica dos Sólidos I

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