min[0.30(55), 0.40(30)] min(16.5, 12.0) 12.0 kpsi
|
|
|
- Lucinda Hall
- 5 years ago
- Views:
Transcription
1 9-10 Table A-0: 100 CD: S ut = 68 kpsi, S = 57 kpsi 100 HR: S ut = 55 kpsi, S = 0 kpsi Cold-rolled properties degrade to hot-rolled properties in the neighborhood of the weld. Table 9-4: all min(0.0 Sut, 0.40 S) min[0.0(55), 0.40(0)] min(16.5, 1.0) 1.0 kpsi for both materials. Eq. (9-): F = 0.707hl all = 0.707(5/16)[()](1.0) = 10.6 kip Ans.
2 9-17 b = d =50 mm, c = 150 mm, h = 5 mm, and allow = 140 MPa. (a) Primar shear, Table 9-1, Case (Note: b and d are interchanged between problem figure and table figure. Note, also, F in kn and in MPa): V F 10.89F A Secondar shear, Table 9-1: J u d b d mm 6 6 J = h J u = 0.707(5)(8.)(10 ) = 94.6(10 ) mm 4 Mr 175F 10 5 x 14.85F J x F F (1) allow 140 F 6.06 kn Ans..1.1 (b) For E7010 from Table 9-6, allow = 1 kpsi = 1(6.89) = 145 MPa 100 HR bar: S ut = 80 MPa, S = 10 MPa 1015 HR support: S ut = 40 MPa, S = 190 MPa Table 9-, E7010 Electrode: S ut = 48 MPa, S = 9 MPa The support controls the design. Table 9-4: allow = min(0.0s ut, 0.40S ) =min[0.0(40), 0.40(190) = min(10, 76) = 76 MPa The allowable load, from Eq. (1) is allow 76 F.9 kn Ans..1.1
3 9- Given, b = in, c = 6 in, d = in, h = 5/16 in, allow = 5 kpsi. Primar shear: Secondar shear: Table 9-1: Maximum shear: V F F A /16 bd Ju in 6 6 J = h J u = 0.707(5/16)10.67 =.57 in 4 Mr 7 F(1) x.970f J.57 x F F allow 5 F 5.41 kip Ans
4 9-45 F = kip, T = 15 kipin. Bending: Table 9-: A = h r = (1/4)(1) = in Torsion: I u = r = (1) =.14 in, I = 0.707(1/4).14 = in 4 V 1.80 kpsi A Mr 1.6 kpsi M I Table 9-1: J u = r = (1) = 6.8 in, J = 0.707(1/4) 6.8 = in Tr 1.5 kpsi T J 1.111
5 Ans M T kpsi.
6 9-5 In the textbook, Fig. Problem 9-5b is a free-bod diagram of the bracket. Forces and moments that act on the welds are equal, but of opposite sense. (a) M = 100(0.66) = 49 lbf in Ans. (b) F = 100 sin 0 = 600 lbf Ans. (c) F x = 100 cos 0 = 109 lbf Ans. (d) From Table 9-, case 6: A 1.414(0.5)(0.5.5) 0.97 in d.5 Iu ( b d) [(0.5).5].9 in 6 6 The second area moment about an axis through G and parallel to z is I 0.707hI (0.5)(.9) in Ans. u (e) Refer to Fig. Problem 9-5b. The shear stress due to F is F A psi The shear stress along the throat due to F x is F x psi A 0.97 The resultant of 1 and is in the throat plane
7 psi The bending of the throat gives Mc 49(1.5) I psi The imum shear stress is psi Ans. (f) Materials: 1018 HR Member: S = kpsi, S ut = 58 kpsi (Table A-0) E6010 Electrode: S = 50 kpsi (Table 9-) n Ss 0.577S 0.577() 1.0 A ns (g) Bending in the attachment near the base. The cross-sectional area is approximatel equal to bh. A1 bh 0.5(.5) 0.65 in Fx 109 x 166 psi A I bd 0.5(.5) 0.60 in c 6 6 At location A, F M A1 I / c psi The von Mises stress is x 648 (166) 91 psi Thus, the factor of safet is, S n 8.18 Ans..91 The clip on the mooring line bears against the side of the 1/-in hole. If the clip fills the hole
8 F psi td 0.5(0.50) S (10 ) n. Ans Further investigation of this situation requires more detail than is included in the task statement. (h) In shear fatigue, the weakest constituent of the weld melt is 1018 HR with S ut = 58 kpsi, Eq. (6-8), p. 8, gives S e 0.504S 0.504(58) 9. kpsi ut Eq. (6-19), p. 87: k a = 14.4(58) = For the size factor estimate, we first emplo Eq. (6-5), p. 89, for the equivalent diameter d hb (.5)(0.5) 0.57 in e Eq. (6-0), p. 88, is used next to find k b k b de Eq.(6-6), p. 90: k c = 0.59 From Eq. (6-18), p. 87, the endurance strength in shear is S se = 0.780(0.940)(0.59)(9.) = 1.6 kpsi From Table 9-5, the shear stress-concentration factor is K f s =.7. The loading is repeatedl-applied 1.57 a m K f s.7.07 kpsi Table 6-7, p. 07: Gerber factor of safet n f, adjusted for shear, with S su = 0.67S ut 1S su a m S se n f 1 1 m Sse Ssua (58).07 (.07)(1.6) Ans (58)(.07) Attachment metal should be checked for bending fatigue.
9 ,F~" (~a).s/tf)c;j C 6rn,cM vjiac) ~ 0- wale CMIJ/, adhl'jivt- hs11.~ ~ Cbh (r'c):t'4i ~I&... u J c~e&,s lo_le-d 6e!&-w. tf'mp(1jvui ;k c1y8(md bocw fe;t..d fo ~sr!oyu F hr pi. ii a) Oc!1YIL F.. Ho'fYllMi Fx d. S /YlJI (r (~A>tt; t?b?et>ami oj p? R OJ ~J(~(f~ av1~1 ~/~~ &.J:l1l) + OJ o.lrrft~.1 up c&~l!fi&v. d p+ o fjj tth ~ (~(VJ1f<.). broc/~/ C{)trJY( H(V-- of 0 /J DhJC ~ /lrjt 1~8tAJU bj If:vw- '< Pr. 1w.L!lJ tj. /YuJPrl 1 b~ /9). lx d,g:a~) hvwv, '. f) I' fy('v1).[a ~,e4' in.;j,a- ci' \". /I-t.,)J A (~A!fr(/t ~ J!,f,.e 0., ' /(;..4 r f/v11(1 '1ft) 'VtA hyo. (M -t/..e~-;-? / r~ YO 1;'<d 6'(Q tim fo -&.1 I ~ru-.f f.ij~ J, ;"- C~{CL ~ijm'j.:;k
Chapter 9. Figure for Probs. 9-1 to Given, b = 50 mm, d = 50 mm, h = 5 mm, τ allow = 140 MPa.
Chapter 9 Figure for Probs. 9-1 to 9-4 9-1 Given, b 50 mm, d 50 mm, h 5 mm, ow 140 MPa. F 0.707 hl ow 0.707(5)[(50)](140)(10 ) 49.5 kn Ans. 9- Given, b in, d in, h 5/16 in, ow 5 kpsi. F 0.707 hl ow 0.707(5/16)[()](5).1
Chapter 9. Figure for Probs. 9-1 to Given, b = 50 mm, d = 50 mm, h = 5 mm, allow = 140 MPa.
Capter 9 Figure for Probs. 9-1 to 9-4 9-1 Given, b = 50 mm, d = 50 mm, = 5 mm, allow = 140 MPa. F = 0.707 l allow = 0.707(5)[(50)](140)(10 ) = 49.5 kn Ans. 9- Given, b = in, d = in, = 5/16 in, allow =
Chapter 9. τ all = min(0.30s ut,0.40s y ) = min[0.30(58), 0.40(32)] = min(17.4, 12.8) = 12.8 kpsi 2(32) (5/16)(4)(2) 2F hl. = 18.1 kpsi Ans. 1.
![Chapter 9. τ all = min(0.30s ut,0.40s y ) = min[0.30(58), 0.40(32)] = min(17.4, 12.8) = 12.8 kpsi 2(32) (5/16)(4)(2) 2F hl. = 18.1 kpsi Ans. 1.](/thumbs/92/108328699.jpg "Chapter 9. τ all = min(0.30s ut,0.40s y ) = min[0.30(58), 0.40(32)] = min(17.4, 12.8) = 12.8 kpsi 2(32) (5/16)(4)(2) 2F hl. = 18.1 kpsi Ans. 1.")
budynas_sm_c09.qxd 01/9/007 18:5 Page 39 Capter 9 9-1 Eq. (9-3: F 0.707lτ 0.707(5/1(4(0 17.7 kip 9- Table 9-: τ all 1.0 kpsi f 14.85 kip/in 14.85(5/1 4.4 kip/in F fl 4.4(4 18.5 kip 9-3 Table A-0: 1018
Tutorial #1 - CivE. 205 Name: I.D:
Tutorial # - CivE. 0 Name: I.D: Eercise : For the Beam below: - Calculate the reactions at the supports and check the equilibrium of point a - Define the points at which there is change in load or beam
Chapter kn m/kg Ans kn m/kg Ans. 187 kn m/kg Ans.
Chapter -1 From Tables A-0, A-1, A-, and A-4c, (a) UNS G1000 HR: S ut = 80 (55) MPa (kpsi), S yt = 10 (0) MPa (kpsi) Ans. (b) SAE 1050 CD: S ut = 690 (100) MPa (kpsi), S yt = 580 (84) MPa (kpsi) Ans. (c)
ENT345 Mechanical Components Design
1) LOAD AND STRESS ANALYSIS i. Principle stress ii. The maximum shear stress iii. The endurance strength of shaft. 1) Problem 3-71 A countershaft carrying two-v belt pulleys is shown in the figure. Pulley
Chapter 10 1" 2 4" 1 = MPa mm. 201 = kpsi (0.105) S sy = 0.45(278.7) = kpsi D = = in. C = D d = = 10.
budynas_sm_ch10.qxd 01/29/2007 18:26 Page 261 Chapter 10 10-1 1" 4" 1" 2 1" 4" 1" 2 10-2 A = Sd m dim( A uscu ) = dim(s) dim(d m ) = kpsi in m dim( A SI ) = dim(s 1 ) dim ( d m ) 1 = MPa mm m A SI = MPa
PROBLEM 5.1 SOLUTION. Reactions: Pb L Pa L. From A to B: 0 < x < a. Pb L Pb L Pb L Pbx L. From B to C: a < x < L Pa L. Pa L. L Pab At section B: M = L
PROBEM 5.1 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the equations of the shear and bending-moment curves. SOUTION Reactions: From A to B: 0 < x < a
ME111 Instructor: Peter Pinsky Class #21 November 13, 2000
Toda s Topics ME Instructor: Peter Pinsk Class # November,. Consider two designs of a lug wrench for an automobile: (a) single ended, (b) double ended. The distance between points A and B is in. and the
WORCESTER POLYTECHNIC INSTITUTE
WORCESTER POLYTECHNIC INSTITUTE MECHANICAL ENGINEERING DEPARTMENT STRESS ANALYSIS ES-2502, C 2012 Lecture 17: 10 February 2012 General information Instructor: Cosme Furlong HL-151 (508) 831-5126 cfurlong@wpi.edu
Determine the resultant internal loadings acting on the cross section at C of the beam shown in Fig. 1 4a.
E X M P L E 1.1 Determine the resultant internal loadings acting on the cross section at of the beam shown in Fig. 1 a. 70 N/m m 6 m Fig. 1 Support Reactions. This problem can be solved in the most direct
Stress Transformation Equations: u = +135 (Fig. a) s x = 80 MPa s y = 0 t xy = 45 MPa. we obtain, cos u + t xy sin 2u. s x = s x + s y.
014 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently 9 7. Determine the normal stress and shear stress acting
Failure from static loading
Failure from static loading Topics Quiz /1/07 Failures from static loading Reading Chapter 5 Homework HW 3 due /1 HW 4 due /8 What is Failure? Failure any change in a machine part which makes it unable
University of Pretoria Department of Mechanical & Aeronautical Engineering MOW 227, 2 nd Semester 2014
Universit of Pretoria Department of Mechanical & Aeronautical Engineering MOW 7, nd Semester 04 Semester Test Date: August, 04 Total: 00 Internal eaminer: Duration: hours Mr. Riaan Meeser Instructions:
Endurance Strength Pg 274
[Pg / 8] Fatigue Analysis Pg 257 The Units used as standard: in, kip, kpsi, sec, hp in, kip, kpsi, sec/min, hp Endurance Strength Pg 274 Fatigue failure occurs when a machine element is sujected to fluctuating
University of Waterloo. Partial notes Part 6 (Welded Joints) Fall 2005
University of Waterloo Department of Mechanical Engineering ME 3 - Mechanical Design 1 artial notes art 6 (Welded Joints) (G. Glinka) Fall 005 1. Introduction to the Static Strength Analysis of Welded
[5] Stress and Strain
![[5] Stress and Strain](/thumbs/95/123344550.jpg "[5] Stress and Strain")
[5] Stress and Strain Page 1 of 34 [5] Stress and Strain [5.1] Internal Stress of Solids [5.2] Design of Simple Connections (will not be covered in class) [5.3] Deformation and Strain [5.4] Hooke s Law
Tuesday, February 11, Chapter 3. Load and Stress Analysis. Dr. Mohammad Suliman Abuhaiba, PE
1 Chapter 3 Load and Stress Analysis 2 Chapter Outline Equilibrium & Free-Body Diagrams Shear Force and Bending Moments in Beams Singularity Functions Stress Cartesian Stress Components Mohr s Circle for
Solution: The moment of inertia for the cross-section is: ANS: ANS: Problem 15.6 The material of the beam in Problem
Problem 15.4 The beam consists of material with modulus of elasticity E 14x10 6 psi and is subjected to couples M 150, 000 in lb at its ends. (a) What is the resulting radius of curvature of the neutral
Stress and Strain ( , 3.14) MAE 316 Strength of Mechanical Components NC State University Department of Mechanical & Aerospace Engineering
(3.8-3.1, 3.14) MAE 316 Strength of Mechanical Components NC State Universit Department of Mechanical & Aerospace Engineering 1 Introduction MAE 316 is a continuation of MAE 314 (solid mechanics) Review
CHAPTER 8 SCREWS, FASTENERS, NONPERMANENT JOINTS
CHAPTER 8 SCREWS, FASTENERS, NONPERMANENT JOINTS This chapter deals with the design and analysis of nonpermanent fasteners such as bolts, power screws, cap screws, setscrews, eys and pins. 8- Standards
Chapter Objectives. Copyright 2011 Pearson Education South Asia Pte Ltd
Chapter Objectives To generalize the procedure by formulating equations that can be plotted so that they describe the internal shear and moment throughout a member. To use the relations between distributed
C100 Cryomodule Vacuum Vessel Structural Analysis An Addendum to JLAB-TN
Introduction C100 Cryomodule Vacuum Vessel Structural Analysis An Addendum to JLAB-TN-07-081 Gary G. Cheng and Edward F. Daly The C100 cryomodule (CM) vacuum vessel structural analysis per ASME Boiler
MEMS Project 2 Assignment. Design of a Shaft to Transmit Torque Between Two Pulleys
MEMS 029 Project 2 Assignment Design of a Shaft to Transmit Torque Between Two Pulleys Date: February 5, 206 Instructor: Dr. Stephen Ludwick Product Definition Shafts are incredibly important in order
PROBLEM 5.1. wl x. M ( Lx x )
w PROE 5.1 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the equations of the shear and bending-moment curves. SOUTION Reactions: 0: 0 0: 0 Free bod diagram
Bending Stress. Sign convention. Centroid of an area
Bending Stress Sign convention The positive shear force and bending moments are as shown in the figure. Centroid of an area Figure 40: Sign convention followed. If the area can be divided into n parts
Mechanics of Materials
Mechanics of Materials 2. Introduction Dr. Rami Zakaria References: 1. Engineering Mechanics: Statics, R.C. Hibbeler, 12 th ed, Pearson 2. Mechanics of Materials: R.C. Hibbeler, 9 th ed, Pearson 3. Mechanics
ME 202 STRENGTH OF MATERIALS SPRING 2014 HOMEWORK 4 SOLUTIONS
ÇANKAYA UNIVERSITY MECHANICAL ENGINEERING DEPARTMENT ME 202 STRENGTH OF MATERIALS SPRING 2014 Due Date: 1 ST Lecture Hour of Week 12 (02 May 2014) Quiz Date: 3 rd Lecture Hour of Week 12 (08 May 2014)
MECHANICS OF MATERIALS Sample Problem 4.2
Sample Problem 4. SOLUTON: Based on the cross section geometry, calculate the location of the section centroid and moment of inertia. ya ( + Y Ad ) A A cast-iron machine part is acted upon by a kn-m couple.
Static Failure (pg 206)
Static Failure (pg 06) All material followed Hookeʹs law which states that strain is proportional to stress applied, until it exceed the proportional limits. It will reach and exceed the elastic limit
Stress Analysis Lecture 3 ME 276 Spring Dr./ Ahmed Mohamed Nagib Elmekawy
Stress Analysis Lecture 3 ME 276 Spring 2017-2018 Dr./ Ahmed Mohamed Nagib Elmekawy Axial Stress 2 Beam under the action of two tensile forces 3 Beam under the action of two tensile forces 4 Shear Stress
Lecture 0. Statics. Module 1. Overview of Mechanics Analysis. IDeALab. Prof. Y.Y.KIM. Solid Mechanics
Lecture 0. Statics Module 1. Overview of Mechanics Analysis Overview of Mechanics Analysis Procedure of Solving Mechanics Problems Objective : Estimate the force required in the flexor muscle Crandall,
Module 11 Design of Joints for Special Loading. Version 2 ME, IIT Kharagpur
Module 11 Design of Joints for Special Loading Version ME, IIT Kharagpur Lesson Design of Eccentrically Loaded Welded Joints Version ME, IIT Kharagpur Instructional Objectives: At the end of this lesson,
Stress Analysis Lecture 4 ME 276 Spring Dr./ Ahmed Mohamed Nagib Elmekawy
Stress Analysis Lecture 4 ME 76 Spring 017-018 Dr./ Ahmed Mohamed Nagib Elmekawy Shear and Moment Diagrams Beam Sign Convention The positive directions are as follows: The internal shear force causes a
MECHANICS OF MATERIALS
CHATR Stress MCHANICS OF MATRIALS and Strain Axial Loading Stress & Strain: Axial Loading Suitability of a structure or machine may depend on the deformations in the structure as well as the stresses induced
COLUMNS: BUCKLING (DIFFERENT ENDS)
COLUMNS: BUCKLING (DIFFERENT ENDS) Buckling of Long Straight Columns Example 4 Slide No. 1 A simple pin-connected truss is loaded and supported as shown in Fig. 1. All members of the truss are WT10 43
Errata Sheet for S. D. Rajan, Introduction to Structural Analysis & Design (1 st Edition) John Wiley & Sons Publication
S D Rajan, Introduction to Structural Analsis & Design ( st Edition) Errata Sheet for S D Rajan, Introduction to Structural Analsis & Design ( st Edition) John Wile & Sons Publication Chapter Page Correction
SOLUTION (17.3) Known: A simply supported steel shaft is connected to an electric motor with a flexible coupling.
SOLUTION (17.3) Known: A simply supported steel shaft is connected to an electric motor with a flexible coupling. Find: Determine the value of the critical speed of rotation for the shaft. Schematic and
Problem " Â F y = 0. ) R A + 2R B + R C = 200 kn ) 2R A + 2R B = 200 kn [using symmetry R A = R C ] ) R A + R B = 100 kn
![Problem  F y = 0. ) R A + 2R B + R C = 200 kn ) 2R A + 2R B = 200 kn [using symmetry R A = R C ] ) R A + R B = 100 kn](/thumbs/88/116790835.jpg "Problem  F y = 0. ) R A + 2R B + R C = 200 kn ) 2R A + 2R B = 200 kn [using symmetry R A = R C ] ) R A + R B = 100 kn")
Problem 0. Three cables are attached as shown. Determine the reactions in the supports. Assume R B as redundant. Also, L AD L CD cos 60 m m. uation of uilibrium: + " Â F y 0 ) R A cos 60 + R B + R C cos
ME 323 FINAL EXAM FALL SEMESTER :00 PM 9:00 PM Dec. 16, 2010
ME 33 FINA EXAM FA SEMESTER 1 7: PM 9: PM Dec. 16, 1 Instructions 1. Begin each problem in the space provided on the eamination sheets. If additional space is required, use the paper provided. Work on
Springs Lecture 3. Extension Springs
Springs Lecture 3 Extension Springs Carry tensile loading Extension springs Max. Normal stress at A due to Bending and Axial loading σ A K A 6D 4 F K A + 3 πd πd 4C 4C C ( C ) r d C Max. torsional stress
Donald P. Shiley School of Engineering ME 328 Machine Design, Spring 2019 Assignment 1 Review Questions
Donald P. Shiley School of Engineering ME 328 Machine Design, Spring 2019 Assignment 1 Review Questions Name: This is assignment is in workbook format, meaning you may fill in the blanks (you do not need
five Mechanics of Materials 1 ARCHITECTURAL STRUCTURES: FORM, BEHAVIOR, AND DESIGN DR. ANNE NICHOLS SUMMER 2017 lecture
ARCHITECTURAL STRUCTURES: FORM, BEHAVIOR, AND DESIGN DR. ANNE NICHOLS SUMMER 2017 lecture five mechanics www.carttalk.com of materials Mechanics of Materials 1 Mechanics of Materials MECHANICS MATERIALS
Revision Term 2. Prof Ahmed Kovacevic
ME 1110 Engineering Practice 1 Engineering Drawing and Design - Lecture 20 Revision Term 2 Prof Ahmed Kovacevic School of Engineering and Mathematical Sciences Room CG25, Phone: 8780, E-Mail: a.kovacevic@city.ac.uk
MECHANICS OF MATERIALS
Third E CHAPTER 2 Stress MECHANICS OF MATERIALS Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf Lecture Notes: J. Walt Oler Texas Tech University and Strain Axial Loading Contents Stress & Strain:
RODS: THERMAL STRESS AND STRESS CONCENTRATION
RODS: HERML SRESS ND SRESS CONCENRION Example 5 rod of length L, cross-sectional area, and modulus of elasticity E, has been placed inside a tube of the same length L, but of cross-sectional area and modulus
Problem d d d B C E D. 0.8d. Additional lecturebook examples 29 ME 323
Problem 9.1 Two beam segments, AC and CD, are connected together at C by a frictionless pin. Segment CD is cantilevered from a rigid support at D, and segment AC has a roller support at A. a) Determine
Strength and Stiffness of Engineering Systems
Numerical SOLUTIONS to Odd-Numbered roblems* of Chapters 1 10 of: Strength and Stiffness of Engineering Systems by: Frederick A. Leckie Dominic J. Dal Bello Solutions Version: 2009.1 Solutions by: D. J.
CHAPTER 2 Failure/Fracture Criterion
(11) CHAPTER 2 Failure/Fracture Criterion (12) Failure (Yield) Criteria for Ductile Materials under Plane Stress Designer engineer: 1- Analysis of loading (for simple geometry using what you learn here
CHAPTER 4. Stresses in Beams
CHAPTER 4 Stresses in Beams Problem 1. A rolled steel joint (RSJ) of -section has top and bottom flanges 150 mm 5 mm and web of size 00 mm 1 mm. t is used as a simply supported beam over a span of 4 m
Homework 6.1 P = 1000 N. δ δ δ. 4 cm 4 cm 4 cm. 10 cm
Homework 6.1 Three thick and wide boards are connected together by two parallel rows of uniformly distributed nails separated by longitude distance δ to form a beam that is subject to constant vertical
MECHANICS OF MATERIALS
Third E CHAPTER 6 Shearing MECHANCS OF MATERALS Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf Lecture Notes: J. Walt Oler Texas Tech University Stresses in Beams and Thin- Walled Members Shearing
APPENDIX A Thickness of Base Metal
APPENDIX A Thickness of Base Metal For uncoated steel sheets, the thickness of the base metal is listed in Table A.1. For galvanized steel sheets, the thickness of the base metal can be obtained by subtracting
Mechanics of Materials II. Chapter III. A review of the fundamental formulation of stress, strain, and deflection
Mechanics of Materials II Chapter III A review of the fundamental formulation of stress, strain, and deflection Outline Introduction Assumtions and limitations Axial loading Torsion of circular shafts
EMA 3702 Mechanics & Materials Science (Mechanics of Materials) Chapter 6 Shearing Stress in Beams & Thin-Walled Members
EMA 3702 Mechanics & Materials Science (Mechanics of Materials) Chapter 6 Shearing Stress in Beams & Thin-Walled Members Beams Bending & Shearing EMA 3702 Mechanics & Materials Science Zhe Cheng (2018)
THREE HOLE SPLICE CLEVIS FOR B52 Standard finishes: ZN, GRN Wt./C 126 Lbs. (57.1 kg)
CLEVIS FITTINGS B169 TWO HOLE SPLICE Wt./C 84 Lbs. (38.1 kg) B168 THREE HOLE SPLICE Wt./C 126 Lbs. (57.1 kg) B167 Wt./C 178 Lbs. (80.7 kg) 3 1 /2 (88.9) B170 TWO HOLE SPLICE Wt./C 123 Lbs. (55.8 kg) 5
ENGINEERING MECHANICS STATIC
Trusses Simple trusses The basic element of a truss is the triangle, three bars joined by pins at their ends, fig. a below, constitutes a rigid frame. The term rigid is used to mean noncollapsible and
ADDENDUM. Revise the ISO title on the Cover to read:
Date of Issue: October 2015 Affected Publication: ANSI/API Technical Report 5C3/ISO 10400:2007, Technical Report on Equations and Calculations or Casing, Tubing, and Line Pipe Used as Casing or Tubing;
MECE 3321 MECHANICS OF SOLIDS CHAPTER 1
MECE 3321 MECHANICS O SOLIDS CHAPTER 1 Samantha Ramirez, MSE WHAT IS MECHANICS O MATERIALS? Rigid Bodies Statics Dynamics Mechanics Deformable Bodies Solids/Mech. Of Materials luids 1 WHAT IS MECHANICS
PROBLEM 7.37 SOLUTION
PROLEM 7.37 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the maimum absolute values of the shear and bending moment. Free bod: Entire beam Σ M = 0: E(6
8.1 Internal Forces in Structural Members
8.1 Internal Forces in Structural Members 8.1 Internal Forces in Structural Members xample 1, page 1 of 4 1. etermine the normal force, shear force, and moment at sections passing through a) and b). 4
WORCESTER POLYTECHNIC INSTITUTE
WORCESTER POLYTECHNIC INSTITUTE MECHANICAL ENGINEERING DEPARTMENT DESIGN OF MACHINE ELEMENTS ME-3320, B 2018 Lecture 04 October 2018 Office hours: Announcements (also in the website of our course) Prof.
MECHANICS OF MATERIALS. Prepared by Engr. John Paul Timola
MECHANICS OF MATERIALS Prepared by Engr. John Paul Timola Mechanics of materials branch of mechanics that studies the internal effects of stress and strain in a solid body. stress is associated with the
Errata (Includes critical corrections only for the 1 st, 2 nd & 3 rd reprints)
Page Number Errata (Includes critical corrections only for the 1 st, 2 nd & 3 rd reprints) Description of Correction Mechanics of Materials, 8e James M. Gere & Barry J. Goodno ISBN: 9781111577735 11 Figure
Spherical Pressure Vessels
Spherical Pressure Vessels Pressure vessels are closed structures containing liquids or gases under essure. Examples include tanks, pipes, essurized cabins, etc. Shell structures : When essure vessels
Presented by: Civil Engineering Academy
Presented by: Civil Engineering Academy Structural Design and Material Properties of Steel Presented by: Civil Engineering Academy Advantages 1. High strength per unit length resulting in smaller dead
Cover sheet and Problem 1
over sheet and Problem nstructions M 33 FNL M FLL SMSTR 0 Time allowed: hours. There are 4 problems, each problem is of equal value. The first problem consists of three smaller sub-problems. egin each
Spring Main Exam / STUDENT NUMBER: I I J.. ll I I I I SURNAME: (FAMILY NAME) OTHER NAMES: < trength of Engineering Materi.a.
Spring 2014- Main Exam / iuts UNIVERSITY OF TECHNOLOGY, SYDNEY STUDENT NUMBER: I I J.. ll I I I I SURNAME: (FAMILY NAME) OTHER NAMES: ''''''''"'.;..,;..:.;..,,,,"_'"'''"'"'"''''"""''"'""'"'''''"'"''"'"'''''"'"'"'''''''"'''''''_'_''"'""''''"''"'"'''''"''''''''""""'"'''"'""'"'""""'"-
Project of a scissor lift platform and concept of an extendable cabin
Project of a scissor lift platform and concept of an extendable cabin Ricardo Duarte Cabrita ricardo.ricab13@gmail.com Instituto Superior Técnico, Lisbon, Portugal May 2016 Abstract This report presents
IDE 110 Mechanics of Materials Spring 2006 Final Examination FOR GRADING ONLY
Spring 2006 Final Examination STUDENT S NAME (please print) STUDENT S SIGNATURE STUDENT NUMBER IDE 110 CLASS SECTION INSTRUCTOR S NAME Do not turn this page until instructed to start. Write your name on
SECTION 7 DESIGN OF COMPRESSION MEMBERS
SECTION 7 DESIGN OF COMPRESSION MEMBERS 1 INTRODUCTION TO COLUMN BUCKLING Introduction Elastic buckling of an ideal column Strength curve for an ideal column Strength of practical column Concepts of effective
Please review the following statement: I certify that I have not given unauthorized aid nor have I received aid in the completion of this exam.
Please review the followg statement: I certify that I have not given unauthorized aid nor have I received aid the completion of this eam. Signature: INSTRUCTIONS Beg each problem the space provided on
Fig Example 8-4. Rotor shaft of a helicopter (combined torsion and axial force).
PEF 309 Fundamentos de Mecânica das Estruturas Eercícios resolvidos etraídos de Mechanics of Materials, th edition, GERE, J.M. e Timoshenko, S.P.,PWS Publishing Company, 997, Boston, USA, p.08-0, 580-583.
Errata (Includes critical corrections only for the 1 st, 2 nd & 3 rd reprints)
Errata (Includes critical corrections only for the 1 st, 2 nd & 3 rd reprints) Mechanics of Materials, 8e James M. Gere & Barry J. Goodno ISBN: 9781111577735 Page Number Description of Correction 15 Example
Beams. Beams are structural members that offer resistance to bending due to applied load
Beams Beams are structural members that offer resistance to bending due to applied load 1 Beams Long prismatic members Non-prismatic sections also possible Each cross-section dimension Length of member
5 G R A TINGS ENGINEERING DESIGN MANUAL. MBG Metal Bar Grating METAL BAR GRATING MANUAL MBG METAL BAR GRATING NAAMM
METAL BAR NAAMM GRATNG MANUAL MBG 534-12 5 G R A TNG NAAMM MBG 534-12 November 4, 2012 METAL BAR GRATNG ENGNEERNG DEGN MANUAL NAAMM MBG 534-12 November 4, 2012 5 G R A TNG MBG Metal Bar Grating A Division
Shafts Introduction. Shafts 509
Shafts 509 C H A P T E R 14 Shafts 1. Introduction.. Material Used for Shafts.. Manufacturing of Shafts. 4. Types of Shafts. 5. Standard Sizes of Transmission Shafts. 6. Stresses in Shafts. 7. Maximum
5. What is the moment of inertia about the x - x axis of the rectangular beam shown?
1 of 5 Continuing Education Course #274 What Every Engineer Should Know About Structures Part D - Bending Strength Of Materials NOTE: The following question was revised on 15 August 2018 1. The moment
CH. 5 TRUSSES BASIC PRINCIPLES TRUSS ANALYSIS. Typical depth-to-span ratios range from 1:10 to 1:20. First: determine loads in various members
CH. 5 TRUSSES BASIC PRINCIPLES Typical depth-to-span ratios range from 1:10 to 1:20 - Flat trusses require less overall depth than pitched trusses Spans: 40-200 Spacing: 10 to 40 on center - Residential
MECHANICS OF MATERIALS Design of a Transmission Shaft
Design of a Transmission Shaft If power is transferred to and from the shaft by gears or sprocket wheels, the shaft is subjected to transverse loading as well as shear loading. Normal stresses due to transverse
March 24, Chapter 4. Deflection and Stiffness. Dr. Mohammad Suliman Abuhaiba, PE
Chapter 4 Deflection and Stiffness 1 2 Chapter Outline Spring Rates Tension, Compression, and Torsion Deflection Due to Bending Beam Deflection Methods Beam Deflections by Superposition Strain Energy Castigliano
Mechanical Design. Design of Shaft
Mechanical Design Design of Shaft Outline Practical information Shaft design Definition of shaft? It is a rotating member, in general, has a circular cross-section and is used to transmit power. The shaft
MECHANICS OF MATERIALS
Third CHTR Stress MCHNICS OF MTRIS Ferdinand. Beer. Russell Johnston, Jr. John T. DeWolf ecture Notes: J. Walt Oler Texas Tech University and Strain xial oading Contents Stress & Strain: xial oading Normal
and F NAME: ME rd Sample Final Exam PROBLEM 1 (25 points) Prob. 1 questions are all or nothing. PROBLEM 1A. (5 points)
ME 270 3 rd Sample inal Exam PROBLEM 1 (25 points) Prob. 1 questions are all or nothing. PROBLEM 1A. (5 points) IND: In your own words, please state Newton s Laws: 1 st Law = 2 nd Law = 3 rd Law = PROBLEM
For sunshades using the Zee blades wind loads are reduced by 10 psf.
C.R. Laurence Co., Inc. 2503 East Vernon Los Angeles, CA 90058 24 July 2009 SUBJ: CR LAURENCE UNIVERSAL SUN SHADES The CRL Universal Aluminum Sun Shades were evaluated in accordance with the 2006 International
VERTICAL ARM AB M B. MAXIMUM STRESSES occur on opposite sides of the vertical arm. MAXIMUM TENSILE STRESS. s t P A M(d 2 2)
54 CHER 8 pplications of lane Stress Combined Loadings he problems for Section 8.5 are to be solved assuming that the structures behave linearl elasticall and that the stresses caused b two or more loads
CONNECTION DESIGN. Connections must be designed at the strength limit state
CONNECTION DESIGN Connections must be designed at the strength limit state Average of the factored force effect at the connection and the force effect in the member at the same point At least 75% of the
I M P O R T A N T S A F E T Y I N S T R U C T I O N S W h e n u s i n g t h i s e l e c t r o n i c d e v i c e, b a s i c p r e c a u t i o n s s h o
I M P O R T A N T S A F E T Y I N S T R U C T I O N S W h e n u s i n g t h i s e l e c t r o n i c d e v i c e, b a s i c p r e c a u t i o n s s h o u l d a l w a y s b e t a k e n, i n c l u d f o l
DESIGN NOTE AND MECHANICAL PROPERTIES PART
COVER PAGE INDEX 1) INDEX PART A SHEET 1~ OF DESIGN NOTE AND MECHANICAL PROPERTIES PART A SHEET ~ OF ) RETRACTABLE OUTRIGGER PART B SHEET 1~7 OF ) TIEBACK ANCHOR PART C SHEET 1~ OF 7 TOTAL 1 SHEETS PROJECT
Solution ME 323 EXAM #2 FALL SEMESTER :00 PM 9:30 PM Nov. 2, 2010
Solution ME 33 EXAM # FALL SEMESTER 1 8: PM 9:3 PM Nov., 1 Instructions 1. Begin each problem in the space provided on the eamination sheets. If additional space is required, use the paper provided. Work
Chapter 3 Trusses. Member CO Free-Body Diagram. The force in CO can be obtained by using section bb. Equations of Equilibrium.
Chapter 3 Trusses Procedure for analysis 1 Free body diagram: make a decision as to how to cut or section the truss through the members where forces are to be determined. 2 Equation of equilibrium: apply
σ = Eα(T T C PROBLEM #1.1 (4 + 4 points, no partial credit)
PROBLEM #1.1 (4 + 4 points, no partial credit A thermal switch consists of a copper bar which under elevation of temperature closes a gap and closes an electrical circuit. The copper bar possesses a length
Solid Mechanics Homework Answers
Name: Date: Solid Mechanics Homework nswers Please show all of your work, including which equations you are using, and circle your final answer. Be sure to include the units in your answers. 1. The yield
Chapter (a) (b) f/(n x) = f/(69 10) = f/690
Chapter -1 (a) 1 10 8 6 4 0 60 70 80 90 100 110 10 130 140 150 160 170 180 190 00 10 (b) f/(n x) = f/(69 10) = f/690 x f fx fx f/(n x) 60 10 700 0.009 70 1 70 4900 0.0015 80 3 40 19 00 0.0043 90 5 450
ES230 STRENGTH OF MATERIALS
ES230 STRENGTH OF MATERIALS Exam 1 Study Guide. Exam 1: Wednesday, February 8 th, in-class Updated 2/5/17 Purpose of this Guide: To thoroughly prepare students for the exact types of problems that will
Introduction to Structural Member Properties
Introduction to Structural Member Properties Structural Member Properties Moment of Inertia (I): a mathematical property of a cross-section (measured in inches 4 or in 4 ) that gives important information
FME461 Engineering Design II
FME461 Engineering Design II Dr.Hussein Jama Hussein.jama@uobi.ac.ke Office 414 Lecture: Mon 8am -10am Tutorial Tue 3pm - 5pm 10/1/2013 1 Semester outline Date Week Topics Reference Reading 9 th Sept 1
Design against fluctuating load
Design against fluctuating load In many applications, the force acting on the spring is not constants but varies in magnitude with time. The valve springs of automotive engine subjected to millions of
MECHANICS OF MATERIALS
Third CHTR Stress MCHNICS OF MTRIS Ferdinand. Beer. Russell Johnston, Jr. John T. DeWolf ecture Notes: J. Walt Oler Texas Tech University and Strain xial oading Contents Stress & Strain: xial oading Normal
Part 1 is to be completed without notes, beam tables or a calculator. DO NOT turn Part 2 over until you have completed and turned in Part 1.
NAME CM 3505 Fall 06 Test 2 Part 1 is to be completed without notes, beam tables or a calculator. Part 2 is to be completed after turning in Part 1. DO NOT turn Part 2 over until you have completed and