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GERE CAP 10-02 SOLUTIONS, Exercícios de Resistência dos materiais

EXERCÍCIOS DO CAP. 10 DO LIVRO MEC. DOS MATERIAS - GERE

Tipologia: Exercícios

2019

Compartilhado em 13/10/2019

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Baixe GERE CAP 10-02 SOLUTIONS e outras Exercícios em PDF para Resistência dos materiais, somente na Docsity! Problem 10.4-2 The propped cantilever beam shown in the figure supports a uniform load of intensity q on the left-hand half of the beam. Find the reactions RA, RB, and MA, and then draw the shear-force and bending-moment diagrams, labeling all critical ordinates. Solution 10.4-2 Propped cantilever beam SECTION 10.4 Method of Superposition 643 Select RB as redundant. EQUILIBRIUM RELEASED STRUCTURE AND FORCE-DISPLACEMENT RELATIONS COMPATIBILITY B  (B)1  (B)2  0 Substitute for (B)1 and (B)2 and solve for RB: OTHER REACTIONS (FROM EQUILIBRIUM) MA  9qL2 128 RA  57qL 128 RB  7qL 128 MA  qL2 8  RBLRA  qL 2  RB SHEAR-FORCE AND BENDING-MOMENT DIAGRAMS Mmax  945qL2 32,768 x1  57L 128 A B MA RA RB q L — 2 L — 2 A B (B)1  PqL4 384EI (B)2  RBL 3 3EI A BL RB q L 2 L 2 x1 RA RB V O  M O Mmax MA Problem 10.4-3 The figure shows a propped cantilever beam ABC having span length L and an overhang of length a. A concentrated load P acts at the end of the overhang. Determine the reactions RA, RB, and MA for this beam. Also, draw the shear-force and bending-moment diagrams, labeling all critical ordinates. A L a B C MA RA RB P Solution 10.4-3 Beam with an overhang 644 CHAPTER 10 Statically Indeterminate Beams Select MA as redundant. EQUILIBRIUM RELEASED STRUCTURE AND FORCE-DISPL. EQS. COMPATIBILITY A  (A)1  (A)2  0 Substitute for (A)1 and (A)2 and solve for MA: MA  Pa 2 RB  1 L (MA  PL  Pa)RA  1 L (MA  Pa) OTHER REACTIONS (FROM EQUILIBRIUM) SHEAR-FORCE AND BENDING-MOMENT DIAGRAMS RB  P 2L (2L  3a)RA  3Pa 2L A B (A)1 (A)2MA L (A)2  MAL 3EI (A)1  PAL 6EI C P a P 3Pa 2L V O  Pa 2 Pa M O  L 3 Problem 10.4-4 Two flat beams AB and CD, lying in horizontal planes, cross at right angles and jointly support a vertical load P at their midpoints (see figure). Before the load P is applied, the beams just touch each other. Both beams are made of the same material and have the same widths. Also, the ends of both beams are simply supported. The lengths of beams AB and CD are LAB and LCD, respectively. What should be the ratio tAB/tCD of the thicknesses of the beams if all four reactions are to be the same? Solution 10.4-4 Two beams supporting a load P P B D C A tAB tCD For all four reactions to be the same, each beam must support one-half of the load P. DEFLECTIONS CD  (P2)L3CD 48EICD AB  (P2)L3AB 48EIAB COMPATIBILITY AB  CD or MOMENT OF INERTIA tAB tCD  LAB LCD ∴ L3AB t3AB  L3CD t3CD ICD  1 12 bt3CDIAB  1 12 bt3AB L3AB IAB  L3CD ICD SECTION 10.4 Method of Superposition 647 Problem 10.4-7 Beam ABC is fixed at support A and rests (at point B) upon the midpoint of beam DE (see the first part of the figure). Thus, beam ABC may be represented as a propped cantilever beam with an overhang BC and a linearly elastic support of stiffness k at point B (see the second part of the figure). The distance from A to B is L  10 ft, the distance from B to C is L /2  5 ft, and the length of beam DE is L  10 ft. Both beams have the same flexural rigidity EI. A concentrated load P  1700 lb acts at the free end of beam ABC. Determine the reactions RA, RB, and MA for beam ABC. Also, draw the shear-force and bending-moment diagrams for beam ABC, labeling all critical ordinates. Solution 10.4-7 Beam with spring support D E A B C P = 1700 lb L = 10 ft = 5 ft k MA RA RB B C P A L 2 — Select RB as redundant. EQUILIBRIUM RA  RB  P MA  RBL  3PL2 RELEASED STRUCTURE AND FORCE-DISPL. EQS. COMPATIBILITY Beam DE: RB  28P 17 7PL3 12 EI  RBL 3 3 EI  RB L 3 48EI k  48 EI L3 B  (B)1  (B)2  RB k OTHER REACTIONS (FROM EQUILIBRIUM) NUMERICAL VALUES P  1700 lb L  10 ft  120 in. RA  1100 lb RB  2800 lb MA  30,000 lb-in. SHEAR-FORCE AND BENDING-MOMENT DIAGRAMS  27.27 in. x1  300 11 in. MA  5PL 34 RA  11P 17 A B (B)1  7PL3 12EI (B)2  RBL3 3EI A L 2 L RB C P 1100 1700 V (LB) O  x1 M (LB  IN.) O 102,000 30,000 } Problem 10.4-8 The beam ABC shown in the figure has flexural rigidity EI  4.0 MN·m2. When the loads are applied to the beam, the support at B settles vertically downward through a distance of 6.0 mm. Calculate the reaction RB at support B. Solution 10.4-8 Overhanging beam with support settlement 648 CHAPTER 10 Statically Indeterminate Beams A CB 6 kN/m 3 kN 3 m 1 m 1 m RB6 mm settlement Select RB as redundant.   settlement of support B RELEASED STRUCTURE AND FORCE-DISPL. EQS. COMPATIBILITY B  (B)1  (B)2  (B)3   Substitute for (B)1, (B)2, and (B)3 and solve for RB: NUMERICAL VALUES q  6.0 kN/m P  3.0 kN   6.0 mm L  4.0 m EI  4.0 MN  m2 SUBSTITUTE INTO THE EQUATION FOR RB RB  7.11 kN RB  1 2048 ¢351qL  2816P  6144 EI¢ L3 ≤ A B (B)1  117qL4 2048EI (B)2  11PL3 24EI A B RB q 3L 4 L 4 L 4 P C C C (B)3  RBL 3 3EIA B Problem 10.4-9 A beam AB is cantilevered from a wall at one end and held by a tie rod at the other end (see figure). The beam is an S 6  12.5 I-beam section with length L  6 ft. The tie rod has a diameter of 1/4 inch and length H  3 ft. Both members are made of steel with E  30  106 psi. A uniform load of intensity q  200 lb/ft acts along the length of the beam. Before the load q is applied, the tie rod just meets the end of the cable. (a) Determine the tensile force T in the tie rod due to the uniform load q. (b) Draw the shear-force and bending-moment diagrams for the beam, labeling all critical ordinates. A L = 6 ft B C q = 200 lb/ft H = 3 ft S 6  12.5 in. tie rod 1— 4 Solution 10.4-9 Beam supported by a tie rod SECTION 10.4 Method of Superposition 649 Select the force T in the tie rod as redundant. RELEASED STRUCTURE AND FORCE-DISPLACEMENT RELATIONS COMPATIBILITY (B)1  (B)2  (B)3 or T  3qAL4 8 AL3  24 IH qL4 8 EI  TL3 3 EI  TH EA NUMERICAL VALUES q  200 lb/ft L  6 ft H  3 ft E  30  106 psi Beam: S 6  12.5 I  22.1 in.4 Tie Rod: d  0.25 in. A  0.04909 in.2 Substitute: T  398 lb SHEAR-FORCE AND BENDING-MOMENT DIAGRAMS RA  qL  T  802 lb  14,530 lb-in. x1  23.9 in. x2  24.2 in. MA  qL2 2  TL A A B B B L T T q (B)1  qL4 8EI (B)2  TL3 3EI (B)3  TH EAH C A B L q MA RA T x1 V (lb) M (lb-in.) O O x2 398 802 4760 14,530 Problem 10.4-10 The figure shows a nonprismatic, propped cantilever beam AB with flexural rigidity 2EI from A to C and EI from C to B. Determine all reactions of the beam due to the uniform load of intensity q. (Hint: Use the results of Problems 9.7-1 and 9.7-2.) A B C MA RA RB q 2EI EI L 2 —L 2 — Problem 10.4-13 A beam AC rests on simple supports at points A and C (see figure). A small gap   0.4 in. exists between the unloaded beam and a support at point B, which is midway between the ends of the beam. The beam has total length 2L  80 in. and flexural rigidity EI  0.4  109 lb-in.2 Plot a graph of the bending moment MB at the midpoint of the beam as a function of the intensity q of the uniform load. Hints: Begin by determining the intensity q0 of the load that will just close the gap. Then determine the corresponding bending moment (MB)0. Next, determine the bending moment MB (in terms of q) for the case where q  q0. Finally, make a statically indeterminate analysis and determine the moment MB (in terms of q) for the case where q  q0. Plot MB (units of lb-in.) versus q (units of lb/in.) with q varying from 0 to 2500 lb/in. 652 CHAPTER 10 Statically Indeterminate Beams A L D L q L RA RB RC RD B C 2L 5 3 5 3 5 1 2 1 2 2 5 2 25 2 25 1 40 2 5 V qL M qL2 2L 5    1 10 1 10 O O A C B RA RC L = 40 in. L = 40 in. q RB  = 0.4 in. LOADING, SHEAR-FORCE, AND BENDING-MOMENT DIAGRAMS Mmax  2 qL2 25 MB  MC   qL2 10 SECTION 10.4 Method of Superposition 653 q0  load required to close the gap   magnitude of gap (MB)0  bending moment when q  q0 CASE 1 q  q0 RA  RC  qL CASE 2 q  q0 (1) (2) CASE 3 q  q0 (statically indeterminate) Select RB as redundant. RELEASED STRUCTURE (B)2  RBL 3 6EI (B)1  5qL4 24EI (MB)0  q0 L 2 2  12EI¢ 5 L2 q0  24 EI¢ 5 L4 B  ¢  5q0 L 4 24 EI MB  qL2 2 B  5 qL4 24 EI COMPATIBILITY B  (B)1  (B)2   or EQUILIBRIUM RA  RC 2RA  2qL  RB  0 NUMERICAL VALUES   0.4 in. L  40 in. EI  0.4  109 lb-in.2 Units: lb, in. From eqs. (1) and (2): q0  300 lbin. (MB)0  240,000 lb-in. For q  q0: MB  800 q (3) For q  q0: MB  300,000  200 q (4) GRAPH OF BENDING MOMENT MB (EQS. 3 AND 4) MB  0 at q  1500 MB  RAL  qL2 2  3EI¢ L2  qL2 8 RA  RC  3qL 8  3EI¢ L3 RB  5qL 4  6 EI¢ L3 5qL4 24 EI  RB L 3 6 EI  ¢ Solution 10.4-13 Beam on a support with a gap A CB RA RC L L q RB  A A RB C C q (B)2 (B)1 O 300,000 200,000 20001000 200,000 100,000 100,000  q0  300 EQ. (3) EQ. (4) MB (lb-in.) (MB)0  240,000 q (lb/in.) MB  0 at q  1500 Problem 10.4-14 A fixed-end beam AB of length L is subjected to a moment M0 acting at the position shown in the figure. (a) Determine all reactions for this beam. (b) Draw shear-force and bending-moment diagrams for the special case in which a  b  L /2. Solution 10.4-14 Fixed-end beam (M0  applied load) 654 CHAPTER 10 Statically Indeterminate Beams A L a b B MBMA M0 RA RB Select RB and MB as redundants. L  a  b EQUILIBRIUM RA  RB MA  MB  RBL  M0 RELEASED STRUCTURE AND FORCE-DISPL. EQS. (B)3  MB L 2 2 EI (uB)3  MB L EI (B)2  RB L 3 3 EI (uB)2  RB L 2 2 EI (B)1  M0 a 2 EI (a  2b)(uB)1  M0 a EI COMPATIBILITY B  (B)1  (B)2  (B)3  0 or 2RBL 3  3MBL 2  3M0a(a  2b) (1) B  (B)1  (B)2  (B)3  0 or RBL 2  2MBL  2M0 a (2) SOLVE EQS. (1) AND (2): FROM EQUILIBRIUM: SPECIAL CASE a  b  L2 MA  MB  M0 4 RA  RB  3M0 2L MA  M0 b L2 (3a  L)RA  6M0 ab L3 MB   M0 a L2 (3b  L)RB   6M0 ab L3 A a b B MBMA M0 RA RB (B)1  M0a 2EI A M0 (B)1 ( B)1  M0a EI ( B)2  RBL 2 2EI ( B)3  MBL EI (B)3  MBL 2 2EI (B)2  RBL 3 3EI ( B)1 B A (B)2 ( B)2 ( B)3 B A B (a  2b) RB MB (B)3 3M0 2L M0 4 M0 4 V L 6  M0 2 M0 2  O O M