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Resolução Projeto de Maquinas Norto Disciplina:

Projeto de Máquinas (/disciplina/projeto-de-maquinas)

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VISUALIZAR ARQUIVO COMPLETO Enviado por

Victor

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PRÉ-VISUALIZAÇÃO

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Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-9-1 PROBLEM 4-9 Statement: A ViseGrip plier-wrench is drawn to scale in Figure P4-3, and for which the forces were analyzed in Problem 3-9, find the stresses in each pin for an assumed clamping force of P = 4000 N in the position shown. The pins are 8-mm dia and are all in double shear. Given: Pin forces as calculated in Problem 3-9: Member 1 F21 7.5 kN F41 5.1 kN Member 2 F12 7.5 kN F32 5.1 kN Member 3 F23 5.1 kN F43 5.1 kN Member 4 F14 5.1 kN F34 5.1 kN Pin diameter d 8 mm Assumptions: Links 3 and 4 are in a toggle position, i.e., the pin that joins links 3 and 4 is in line with the pins that join 1 with 4 and 2 with 3. Solution: See Figure 4-9 and Mathcad file P0409. 1. The FBDs of the assembly and each individual link are shown in Figure 4-9. The dimensions, as scaled from Figure P4-3 in the text, are shown on the link FBDs. 50.0 = a 22.0 = d

21.2 = h F 26.9 = f 32 2.8 = g

28.0 = e 2 F12 P 14F 4 34F F 4 2 1 P P F 55.0 = b

F 39.5 = c

3 F41 43F 23F 129.2° F21 1 F 3 P FIGURE 4-9 Free Body Diagrams for Problem 4-9 2. The cross-sectional area for all pins is the same and is A d2 4 A 50.265 mm2 © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-9-2 3. The pin that joins members 1 and 2 is the most highly stressed while the stress on each of the remaining pins is the same. Since the pins are in double shear, we will divide the pin load by 2 in each case. Pin joining 1 and 2 12 F12 2 A 12 74.6 MPa All other pins 14 F14 2 A 14 50.7 MPa © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-10-1 PROBLEM 4-10 Statement: The over-hung diving board of problem 3-10 is shown in Figure P4-4a. Assume cross-section dimensions of 305 mm x 32 mm. The material has E = 10.3 GPa. Find the largest principal stress at any location in the board when a 100-kg person is standing at the free end. 700 = a 2000 = L R R 1 2 P Given: Weight of person W 100 kgf Board dimensions Distance to support a 0.7 m Length of board L 2 m Cross-section w 305 mm t 32 mm FIGURE 4-10 Assumptions: The weight of the beam is negligible compared to the applied load and so can be ignored. Free Body Diagram for Problem 4-10 Solution: See Figure 4-10 and Mathcad file P0410. 1. From the FBD of the diving board and Figure B-3 (Appendix B), the reactions at the supports are R1 W 1 La

R1 1821 N R2 W L a

R2 2802 N 2. Also from Figure D-3, the maximum bending moment occurs at the right-hand support where, in the FBD above, x = a. Mmax R1 a Mmax 1275 N m 3. The maximum bending stress will occur on the top and bottom surfaces of the board at the section where the maximum bending moment occurs which, in this case, is at x = a. The only stress present on the top or bottom surface of the board is the bending stress x. Therefore, on the top surface where the stress is tensile, x. is the principal stress 1 . Thus, Distance to extreme fiber c t 2 c 16 mm Moment of inertia I w t3 12 I 8.329 105 mm4 Bending stress x Mmax c I x 24.492 MPa Maximum principal stress 1 x 1 24.5 MPa © 2011 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458. MACHINE DESIGN - An Integrated Approach, 4th Ed. 4-11-1 PROBLEM 4-11 Statement: Repeat Problem 4-10 using the loading conditions of Problem 3-11. Assume the board weighs 29 kg and deflects 13.1 cm statically when the person stands on it. Find the largest principal stress at any location in the board when the 100-kg person in Problem 4-10 jumps up 25 cm and lands back on the board. Find the maximum deflection. 700 = a 2000 = L R R 1 2 F i Given: Beam length L 2000 mm Distance to support a 700 mm Mass of person mpers 100 kg Mass of board mboard 29 kg Static deflection st 131 mm Height of jump h 250 mm Cross-section w 305 mm FIGURE 4-11 t 32 mm Free Body Diagram for Problem 4-11 Assumptions: The apparent Young's modulus for fiberglas is E 1.03 104 MPa Solution: See Figure 4-11 and Mathcad file P0411. 1. From Problem 3-11, the dynamic load resulting from the impact of the person with the board isFi 3.056 kN 2. From the FBD of the diving board and Figure B-3(a) (Appendix B), the reactions at the supports are R1 Fi 1 La

R1 5.675 kN R2 Fi L a

R2 8.731 kN 3. Also from Figure D-3(a), the maximum bending moment occurs at the right-hand support where, in the FBD above, x = a. Mmax R1 a Mmax 3.973 kN m 4. The maximum bending stress will occur on the top and bottom surfaces of the board at the section where the maximum bending moment occurs which, in this case, is at x = a. The only stress present on the top or bottom surface of the board is the bending stress x. Therefore, on the top surface where the stress is tensile, x is the principal stress 1 . Thus, Distance to extreme fiber c t 2 c 16 mm Moment of inertia I w t3 12 I 8.329 105 mm4 Bending stress x Mmax c I x 76.322 MPa Maximum principal stress 1 x 1 76.3 MPa 5. Calculate the

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