BTECH MECHANICAL PRINCIPLES AND APPLICATIONS. Level 3 Unit 5
|
|
- Opal York
- 6 years ago
- Views:
Transcription
1 BTECH MECHANICAL PRINCIPLES AND APPLICATIONS Level 3 Unit 5
2 FORCES AS VECTORS Vectors have a magnitude (amount) and a direction. Forces are vectors
3 FORCES AS VECTORS (2 FORCES) Forces F1 and F2 are in different F1 directions They are NOT in equilibrium F2
4 FORCES AS VECTORS (2 FORCES) The two forces can be drawn like this. (In the F2 correct direction and the lengths should be drawn to F1 scale to represent the magnitude of the forces)
5 FORCES AS VECTORS (2 FORCES) If the two forces do not F2 meet, the system is not in F1 equilibrium
6 FORCES AS VECTORS (2 FORCES) If a third force (FE) was added in the way shown the three would be in equilibrium (They are all joined up following each other, The force system is balanced) F1 F E F2 This force is called the EQUILIBRANT
7 FORCES AS VECTORS (2 FORCES) If the line joining the two forces is in the opposite direction to F2 the equilibrant it is the F1 RESULTANT of the two forces The forces area not in equilibrium and the resultant shows the direction and magnitude of the combination of the two forces FR
8 FORCES AS VECTORS (3 FORCES) F3 F2 FE F3 F1 F1 F2
9 FORCES AS VECTORS (3 FORCES) F3 F2 FR F3 F1 F1 F2
10 FORCES AS VECTORS ( 3 FORCES) 2 ND EXAMPLE F3 50 o F2 F1 Equilibrant FE F1 Resultant F3 F2 50 o FR 50 o F1 F3 F2
11 Forces on a flat F2 = 4N FORCES AS VECTORS rectangular plate 40cm 60cm 50 o F3 = 12N F1 = 10N Equilibrant Resultant FE F1 FR 10N 50 o 50 o 4N 4N 12N 12N
12 FINDING FE IDENTIFY THE DIRECTION AND MAGNITUDE OF THE FORCES THEN CONSTRUCT A VECTOR DIAGRAM F2 = 4N 60cm 40cm 50 o F1 = 10 N F3 =12 N
13 DRAW TO SCALE TO FIND THE MAGNITUDE AND DIRECTION OF FE (EQUILIBRANT) FE 50 o 10N 4N 12N FE = 22N (Measured)
14 DRAW IT IN THE OPPOSITE DIRECTION TO FIND THE MAGNITUDE AND DIRECTION OF RESULTANT FORCE FR 4N 50 o 10N 12N FE = 22N (Measured)
15 FINDING THE POSITION OF THE EQUILIBRANT (FE) 60cm 4N 22N F1 40cm 10N 12N 4N 12N 50 o
16 Put FE where you think it should be to balance the other forces Clockwise = Anticlockwise 4N 22N x X = 4N x 40cm + 10N x N x 0 60cm 40cm 10N = 160Ncm X = = 7.27 cm 22N 12N x 50 o A The 10N and the 12N pass through the pivot A so the turning moment = 0
17 B TECH Question example P1 The diagram shows a uniform rectangular plate supported in a vertical plane by forces acting at the three corners of the plate. The plate is 4m x 3m and has a mass of 200kg 2.6kN a) Calculate the magnitude and direction of the resultant force 1.4kN A 4m 35 o b) Show the magnitude and direction of the equilibrant force 3m 130 o c) Calculate the position of the resultant force with respect to the corner A (ie. Use A as a pivot) 1.4 kn
18 Weight of plate = 200Kg 1.4kN A 4m 35 o 2.6kN x 9,81 = 1.96kN 3m (acting from 130 o the centre of gravity of the 1.96 kn 1.4 kn uniform plate
19 VECTOR DIAGRAM WITH RESULTANT 2.2kN 1.4kN 1.96kN 2.6kN This shows a) the magnitude and direction of the resultant 1.4kN
20 VECTOR DIAGRAM WITH EQUILIBRANT 2.2kN 1.4kN 1.96kN 2.6kN This shows b) the magnitude and direction of the equilibrant 1.4kN
21 Resolve the diagonal forces 2.6kN and 1.4kN into vertical and horizontal components V1, H1 and V2 (H2 not needed) V2 = 2.6xsin 35 = 1.49kN V1 = 1.4xsin40 = 0.90kN H1 1.4 x cos40 = 1.07kN For explanation Click here 2.2kN A 1.4kN 3m x 1.96 kn 4m V2 V1 40 o 35 o 2.6kN 1.4 kn H2 not needed, it passes through A H
22 Resolve turning moments Clockwise = Anticlockwise 1.96 x 2 + V2 x 4 + V1 x 4 + H1 x 3 1.4kN 2.2kN 3m A x 1.49kN V2 35 o 2.6kN H2 not needed, it passes through A 2.2 x X 1.96 kn 4m 40 o H1 1.07kN 1.4 kn V1 0.9kN
23 Resolve turning moments Clockwise = Anticlockwise 1.96 x x x x x X X = X = X = 1.65X = = 0.75m 1.4kN 2.2kN A x 1.96 kn 4m 1.49kN V2 V1 0.9kN 35 o 40 o 2.6kN 1.4 kn H2 not needed, it passes through A H1 1.07kN
24 RESOLVING FORCES 2000 Newtons
25 Weight suspended by two ropes Draw the perpendicular Identify the angles between the forces A and B and the perpendicular A B Draw the triangle using the angles 35 o 55 o 20 o 70 o 55 o A 2000 N 105 o B The length of the sides of the triangle represent the magnitude of the forces NOT the length of rope 20 o 2000 Newtons
26 USING THE SINE RULE (IF YOU KNOW THE ANGLES) a/sin A = b/sin B = c/sin C angle A = 20 o angle B = 55 o (opposites to sides a & b) Angle C = 105 o and side c represents 2000N 2000 N (c) 55 o 20 o a 105 o b
27 USING THE SINE RULE (IF YOU KNOW THE ANGLES) a/sin A = c/sin C therefore a/sin 20 o = 2000/sin105 o a = 2000 x sin 20 o /sin105 o 55 o a N 105 o b/sin B = c/sin C therefore b/sin 55 o = 2000/sin105 o a = 2000 x sin 55 o /sin105 o 2000 N (c) 20 o b N
28 USING THE COSINE RULE ( IF YOU KNOW ONE ANGLE AND TWO SIDES) F3 F2 = 60N 70 o F1 = 30N
29 USING THE COSINE RULE ( IF YOU KNOW ONE ANGLE AND TWO SIDES) A 2 = B 2 + C 2-2BCcosA (F3) 2 = x60x30x cos110 o = 75.7N F3 (A) F2 = 60N (C) A =110 o 70 o F1 = 30N (B)
30 VERTICAL AND HORIZONTAL COMPONENTS OF FORCES Sketch the diagram F v F F v can be drawn at the other end of the sketch θ F H
31 VERTICAL AND HORIZONTAL COMPONENTS OF FORCES Sketch the diagram sin θ = F v /F F v F F v F.sin θ = F v θ F H cos θ = F H /F back F.cos θ = F H
32 RESTORING FORCE OF TWO FORCES F1(55N) F3 is the restoring force of F1 and F2 F3 Can be drawn to scale 25 o 25 o 70 o F2 (25N) 74.8N 70 o
33 RESTORING FORCE OF TWO FORCES F1(55N) F3 is the restoring force of F1 and F2 70 o F2 (25N) F3 Can be solved by resolving the horizontal and vertical components of F1 and F2 25 o
34 RESTORING FORCE OF TWO FORCES F1(55N) F3 F1v = F1.sin70 o 55sin70 o = 51.68N 25 o 70 o F2 (25N) F1h = F1.cos70 o 55cos70 o = 18.81N
35 RESTORING FORCE OF TWO FORCES F1(55N) F3 F2v = F2.sin25 o 25sin25 o = 10.57N 25 o 70 o F2 (25N) F2h = F2.cos25 o 25cos25 o = 22.66N
36 RESTORING FORCE OF TWO FORCES F1(55N) F3 F3v = F1v + F2v = 62.25N 25 o 70 o F2 (25N) F3h = F1h +F2h = 41.47N
37 RESTORING FORCE OF TWO FORCES F3 (F3) 2 = N (F3) 2 = N F3 = 74.80N
38 RESULTANT OF TWO FORCES Tan θ = opposite/adjacent F3 Tan θ = 62.25/41.47 θ 62.25N Tan θ = 1.5 θ = o 41.47N Direction of F3 = = o
39 MOMENTS OF FORCE 4m 2m 2N 4N Total Anticlockwise moments = Total Clockwise moments 8Nm 8Nm
40 4m MOMENTS OF FORCE 3m 2m 2N 2N Total Anticlockwise moments = Total Clockwise moments 8Nm + 4Nm = 12Nm = 12 Nm 4N
41 4m MOMENTS OF FORCE 3m 2m 2N 2N Total Anticlockwise moments = Total Clockwise moments 8Nm + 4Nm = 12Nm = 12 Nm 4N
42 4m MOMENTS OF FORCE 3m 2m 2N 2N Total Anticlockwise moments = Total Clockwise moments 8Nm + 4Nm = 12Nm = 12 Nm 4N
43 FORCE ON A AND B 10m A 1m 4m 3m 2m B 2N 2N Take A as the pivot 4N Anticlockwise ( B x 10) = (2 x1) + (2 x 5) + (4 x 8) = 44 Nm Force on B = = 4.4N Total downward force = 8 N Force on A = 3.6 N Check this out using B as the pivot
44 Check this out using B as the pivot FORCE ON A AND B 10m 4 N/m uniformly distributed load A 1m 4m 2.5m 2.5m B 2N Take A as the pivot 2N 20 N (UDL) Anticlockwise ( B x 10) = clockwise (2 x1) + (2 x 5) + (20 x 7.5) = 162 Nm Force on B = = 16.2 N Total downward force = 24 N Force on A = 7.8 N UDL 4N/m x length 5m acting from centre of UDL
45 B TECH Question example 6kN 4kN 4kN/m (uniform distributed load) A 2m 3m 5 m B P2 Calculate the support reactions A and B for the simply supported beam in the diagram
46 B TECH Question example solution 6kN 4kN 4kN/m (uniform distributed load) A 2m 3m 40kN 5 m B Uniform load is 4kN/m. total UDL = 4 x 10 = 40kN ( acting from centre of beam)
47 B TECH Question example solution 6kN 4kN 4kN/m (uniform distributed load) A 2m Anticlockwise B x 10 knm B = kNm 3m 40kN 5 m Clockwise moments 6 x 2 = 12kNm 4 x 5 = 20 knm 40 x 5 = 200 knm total = 232kNm B
48 B TECH Question example solution 6kN 4kN 4kN/m (uniform distributed load) A 2m Total downward force = 50kN 40kN 5 m Total upward force A + B = 50kN A = 50kN B A = = 26.8kN
49 B TECH Question example solution 6kN 4kN 4kN/m (uniform distributed load) A 2m Check using B as the pivot 3m 40kN 5 m CW = A x10 ACW = 4 x 5 = 20kNm 40 x 5 = 200kNm 6 x 8 = 48kNm = 268kNm A = 26.8kN B
50 TENSILE STRESS AND STRAIN Stress Strain hardening Necking Ultimate tensile strength Yield strength Fracture Y X (Stress) (Strain) Strain
51 TENSILE STRESS (σ) CROSS SECTIONAL AREA ( πr 2 ) FORCE FORCE Stress = Force Cross sectional area Force direction is perpendicular to cross sectional area
52 TENSILE STRAIN Lo (original length) Increase in length L Young s Modulus Strain = L Lo Stress Strain
53 SHEAR STRESS (τ) Force Force is parallel to cross sectional area of shear Shear stress = Force cross sectional area of shear
54 SHEAR STRAIN Force Shear strain = Change in length original length l l
55 SHEAR STRESS Force Shear Modulus Shear Stress shear Strain
56 C 20kN A B 20kN P3 The diagram shows a shackle joint subjected to a tensile load. The connecting rods A and B are made from steel and the pin c is made from brass. Young s modulus is 210 GPa for steel and 100GPa foe brass. The shear modulus for steel is 140GPa and the shear modulus for brass is 70GPa. The smallest diameter for the connecting rods A and B is 20mm and the diameter of the pin C is 15 mm. a) Calculate the maximum direct stress in the connecting rods b) Calculate the maximum direct strain in the connecting rods c) Calculate the change in length of a 500mm length of connecting rod. d) Calculate the shear stress in the pin e) Calculate the shear strain in the pin
57 C 20kN A B 20kN Maximum direct stress is on the smallest connecting rod diameter which is 20mm (.02m), radius is 10mm (.01m). Cross sectional area = πr 2 = π x.01 2 = 3.14 x 10-4 m 2. Maximum stress = 20x x 10-4 = 6.4 x 10 7 N/m 2 (Pa) Young s modulus for steel = stress strain = 210GPa = 2.1 x10 11 Pa strain = stress Young s modulus Strain = 6.4 x x10 11 =3 x 10-4 m/m
58 C 20kN A B 20kN Maximum direct stress is on the smallest connecting rod diameter which is 20mm (.02m), radius is 10mm (.01m). Cross sectional area = πr 2 = π x.01 2 = 3.14 x 10-4 m 2. strain = elongation original length elongation =strain x original length = 1.5x10-4 m = 0.15mm = 3 x 10-4 m/m x 0.5
59 C 20kN A B 20kN Shear stress in pin Force area = 20kN cross sectional area of pin (π x ) = 20x x10-4 m 2 = 1.13 x10 8 Pa Shear modulus for brass = 7 x Pa. Strain = stress modulus strain = 1.13 x x =
60 F F = 8kN 70 o In the diagram the diameter in the of the bolt shown for the angle is 12mm. It is made from a material with a tensile strength of 500MPa and a shear strength of 300 MPa a) Determine the operational factor of safety in tension. b) Determine the operational factor of safety in shear.
61 Direct force F1 F F = 8kN 70 o Shear force F2 F2 8kN Sin70 o = F1 8kN F1 = 8kN x Sin70 o 70 o F1 = 7.5 kn F1 Cos70 o = F1 8kN Cos70 o F2 = 8kN x F2 = 2.7 kn
62 F F = 8kN 70 o Operational factor of safety = Tensile strength working stress Cross sectional area of the bolt = πr 2 = π x (.006) 2 = 1.13 x10-4 m 2
63 F F = 8kN 70 o Operational factor of safety = Tensile strength working stress Tensile stress = F area 7.5 x x x 10 7 Pa
64 F F = 8kN 70 o Operational factor of safety = Tensile strength working stress Shear stress = F area 2.7 x x x 10 7 Pa
65 F F = 8kN 70 o Operational factor of safety = Tensile strength working stress operational factor of safety in tension. 500 x 10 6 Pa 6.6 x 10 7 Pa = 7.6
66 F F = 8kN 70 o Operational factor of safety = Tensile strength working stress operational factor of safety in shear. 300 x 10 6 Pa 2.4 x 10 7 Pa = 12.5
8-5 Conjugate-Beam method. 8-5 Conjugate-Beam method. 8-5 Conjugate-Beam method. 8-5 Conjugate-Beam method
The basis for the method comes from the similarity of eqn.1 &. to eqn 8. & 8. To show this similarity, we can write these eqn as shown dv dx w d θ M dx d M w dx d v M dx Here the shear V compares with
More informationOUTCOME 1 - TUTORIAL 3 BENDING MOMENTS. You should judge your progress by completing the self assessment exercises. CONTENTS
Unit 2: Unit code: QCF Level: 4 Credit value: 15 Engineering Science L/601/1404 OUTCOME 1 - TUTORIAL 3 BENDING MOMENTS 1. Be able to determine the behavioural characteristics of elements of static engineering
More information[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
More informationBE Semester- I ( ) Question Bank (MECHANICS OF SOLIDS)
BE Semester- I ( ) Question Bank (MECHANICS OF SOLIDS) All questions carry equal marks(10 marks) Q.1 (a) Write the SI units of following quantities and also mention whether it is scalar or vector: (i)
More informationIntroduction. 1.1 Introduction. 1.2 Trigonometrical definitions
Introduction 1.1 Introduction Stress analysis is an important part of engineering science, as failure of most engineering components is usually due to stress. The component under a stress investigation
More informationD : SOLID MECHANICS. Q. 1 Q. 9 carry one mark each. Q.1 Find the force (in kn) in the member BH of the truss shown.
D : SOLID MECHANICS Q. 1 Q. 9 carry one mark each. Q.1 Find the force (in kn) in the member BH of the truss shown. Q.2 Consider the forces of magnitude F acting on the sides of the regular hexagon having
More informationENGINEERING COUNCIL DIPLOMA LEVEL MECHANICS OF SOLIDS D209 TUTORIAL 3 - SHEAR FORCE AND BENDING MOMENTS IN BEAMS
ENGINEERING COUNCIL DIPLOMA LEVEL MECHANICS OF SOLIDS D209 TUTORIAL 3 - SHEAR FORCE AND BENDING MOMENTS IN BEAMS You should judge your progress by completing the self assessment exercises. On completion
More informationName :. Roll No. :... Invigilator s Signature :.. CS/B.TECH (CE-NEW)/SEM-3/CE-301/ SOLID MECHANICS
Name :. Roll No. :..... Invigilator s Signature :.. 2011 SOLID MECHANICS Time Allotted : 3 Hours Full Marks : 70 The figures in the margin indicate full marks. Candidates are required to give their answers
More informationMAAE 2202 A. Come to the PASS workshop with your mock exam complete. During the workshop you can work with other students to review your work.
It is most beneficial to you to write this mock final exam UNDER EXAM CONDITIONS. This means: Complete the exam in 3 hours. Work on your own. Keep your textbook closed. Attempt every question. After the
More informationMECHANICS OF MATERIALS
Third E CHAPTER 1 Introduction MECHANICS OF MATERIALS Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf Lecture Notes: J. Walt Oler Texas Tech University Concept of Stress Contents Concept of Stress
More informationQ1. The figure below shows an apparatus used to locate the centre of gravity of a non-uniform metal rod.
PhysicsAndMathsTutor.com 1 Q1. The figure below shows an apparatus used to locate the centre of gravity of a non-uniform metal rod. The rod is supported horizontally by two wires, P and Q and is in equilibrium.
More informationMechanics of Solids. Mechanics Of Solids. Suraj kr. Ray Department of Civil Engineering
Mechanics Of Solids Suraj kr. Ray (surajjj2445@gmail.com) Department of Civil Engineering 1 Mechanics of Solids is a branch of applied mechanics that deals with the behaviour of solid bodies subjected
More informationISHIK UNIVERSITY DEPARTMENT OF MECHATRONICS ENGINEERING
ISHIK UNIVERSITY DEPARTMENT OF MECHATRONICS ENGINEERING QUESTION BANK FOR THE MECHANICS OF MATERIALS-I 1. A rod 150 cm long and of diameter 2.0 cm is subjected to an axial pull of 20 kn. If the modulus
More informationIDE 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
More information2012 MECHANICS OF SOLIDS
R10 SET - 1 II B.Tech II Semester, Regular Examinations, April 2012 MECHANICS OF SOLIDS (Com. to ME, AME, MM) Time: 3 hours Max. Marks: 75 Answer any FIVE Questions All Questions carry Equal Marks ~~~~~~~~~~~~~~~~~~~~~~
More information2008 FXA THREE FORCES IN EQUILIBRIUM 1. Candidates should be able to : TRIANGLE OF FORCES RULE
THREE ORCES IN EQUILIBRIUM 1 Candidates should be able to : TRIANGLE O ORCES RULE Draw and use a triangle of forces to represent the equilibrium of three forces acting at a point in an object. State that
More informationPES Institute of Technology
PES Institute of Technology Bangalore south campus, Bangalore-5460100 Department of Mechanical Engineering Faculty name : Madhu M Date: 29/06/2012 SEM : 3 rd A SEC Subject : MECHANICS OF MATERIALS Subject
More informationR13. II B. Tech I Semester Regular Examinations, Jan MECHANICS OF SOLIDS (Com. to ME, AME, AE, MTE) PART-A
SET - 1 II B. Tech I Semester Regular Examinations, Jan - 2015 MECHANICS OF SOLIDS (Com. to ME, AME, AE, MTE) Time: 3 hours Max. Marks: 70 Note: 1. Question Paper consists of two parts (Part-A and Part-B)
More informationDetermine 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
More informationSTRESS, STRAIN AND DEFORMATION OF SOLIDS
VELAMMAL COLLEGE OF ENGINEERING AND TECHNOLOGY, MADURAI 625009 DEPARTMENT OF CIVIL ENGINEERING CE8301 STRENGTH OF MATERIALS I -------------------------------------------------------------------------------------------------------------------------------
More information1. A pure shear deformation is shown. The volume is unchanged. What is the strain tensor.
Elasticity Homework Problems 2014 Section 1. The Strain Tensor. 1. A pure shear deformation is shown. The volume is unchanged. What is the strain tensor. 2. Given a steel bar compressed with a deformation
More informationPDDC 1 st Semester Civil Engineering Department Assignments of Mechanics of Solids [ ] Introduction, Fundamentals of Statics
Page1 PDDC 1 st Semester Civil Engineering Department Assignments of Mechanics of Solids [2910601] Introduction, Fundamentals of Statics 1. Differentiate between Scalar and Vector quantity. Write S.I.
More informationM201 assessment Moments
Do the questions as a test circle questions you cannot answer Red 1) A light see-saw is 10 m long with the pivot 3 m from the left. a) A 4 kg weight is placed on the left-hand end of the see-saw. Write
More informationSTRENGTH OF MATERIALS-I. Unit-1. Simple stresses and strains
STRENGTH OF MATERIALS-I Unit-1 Simple stresses and strains 1. What is the Principle of surveying 2. Define Magnetic, True & Arbitrary Meridians. 3. Mention different types of chains 4. Differentiate between
More informationSN QUESTION YEAR MARK 1. State and prove the relationship between shearing stress and rate of change of bending moment at a section in a loaded beam.
ALPHA COLLEGE OF ENGINEERING AND TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING MECHANICS OF SOLIDS (21000) ASSIGNMENT 1 SIMPLE STRESSES AND STRAINS SN QUESTION YEAR MARK 1 State and prove the relationship
More informationCIVIL DEPARTMENT MECHANICS OF STRUCTURES- ASSIGNMENT NO 1. Brach: CE YEAR:
MECHANICS OF STRUCTURES- ASSIGNMENT NO 1 SEMESTER: V 1) Find the least moment of Inertia about the centroidal axes X-X and Y-Y of an unequal angle section 125 mm 75 mm 10 mm as shown in figure 2) Determine
More informationUNIVERSITY OF SASKATCHEWAN ME MECHANICS OF MATERIALS I FINAL EXAM DECEMBER 13, 2008 Professor A. Dolovich
UNIVERSITY OF SASKATCHEWAN ME 313.3 MECHANICS OF MATERIALS I FINAL EXAM DECEMBER 13, 2008 Professor A. Dolovich A CLOSED BOOK EXAMINATION TIME: 3 HOURS For Marker s Use Only LAST NAME (printed): FIRST
More informationNORMAL STRESS. The simplest form of stress is normal stress/direct stress, which is the stress perpendicular to the surface on which it acts.
NORMAL STRESS The simplest form of stress is normal stress/direct stress, which is the stress perpendicular to the surface on which it acts. σ = force/area = P/A where σ = the normal stress P = the centric
More informationAPPLIED MECHANICS I Resultant of Concurrent Forces Consider a body acted upon by co-planar forces as shown in Fig 1.1(a).
PPLIED MECHNICS I 1. Introduction to Mechanics Mechanics is a science that describes and predicts the conditions of rest or motion of bodies under the action of forces. It is divided into three parts 1.
More informationStrength of Materials (15CV 32)
Strength of Materials (15CV 32) Module 1 : Simple Stresses and Strains Dr. H. Ananthan, Professor, VVIET,MYSURU 8/21/2017 Introduction, Definition and concept and of stress and strain. Hooke s law, Stress-Strain
More information9 MECHANICAL PROPERTIES OF SOLIDS
9 MECHANICAL PROPERTIES OF SOLIDS Deforming force Deforming force is the force which changes the shape or size of a body. Restoring force Restoring force is the internal force developed inside the body
More informationSamantha Ramirez, MSE. Stress. The intensity of the internal force acting on a specific plane (area) passing through a point. F 2
Samantha Ramirez, MSE Stress The intensity of the internal force acting on a specific plane (area) passing through a point. Δ ΔA Δ z Δ 1 2 ΔA Δ x Δ y ΔA is an infinitesimal size area with a uniform force
More informationBending 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
More informationFree Body Diagram: Solution: The maximum load which can be safely supported by EACH of the support members is: ANS: A =0.217 in 2
Problem 10.9 The angle β of the system in Problem 10.8 is 60. The bars are made of a material that will safely support a tensile normal stress of 8 ksi. Based on this criterion, if you want to design the
More informationCH. 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
More informationSub. Code:
Important Instructions to examiners: ) The answers should be examined by key words and not as word-to-word as given in the model answer scheme. ) The model answer and the answer written by candidate may
More informationQUESTION BANK SEMESTER: III SUBJECT NAME: MECHANICS OF SOLIDS
QUESTION BANK SEMESTER: III SUBJECT NAME: MECHANICS OF SOLIDS UNIT 1- STRESS AND STRAIN PART A (2 Marks) 1. Define longitudinal strain and lateral strain. 2. State Hooke s law. 3. Define modular ratio,
More informationUnit Workbook 1 Level 4 ENG U8 Mechanical Principles 2018 UniCourse Ltd. All Rights Reserved. Sample
Pearson BTEC Levels 4 Higher Nationals in Engineering (RQF) Unit 8: Mechanical Principles Unit Workbook 1 in a series of 4 for this unit Learning Outcome 1 Static Mechanical Systems Page 1 of 23 1.1 Shafts
More informationEngineering Mechanics: Statics in SI Units, 12e
Engineering Mechanics: Statics in SI Units, 12e 5 Equilibrium of a Rigid Body Chapter Objectives Develop the equations of equilibrium for a rigid body Concept of the free-body diagram for a rigid body
More informationboth an analytical approach and the pole method, determine: (a) the direction of the
Quantitative Problems Problem 4-3 Figure 4-45 shows the state of stress at a point within a soil deposit. Using both an analytical approach and the pole method, determine: (a) the direction of the principal
More informationStress Strain Elasticity Modulus Young s Modulus Shear Modulus Bulk Modulus. Case study
Stress Strain Elasticity Modulus Young s Modulus Shear Modulus Bulk Modulus Case study 2 In field of Physics, it explains how an object deforms under an applied force Real rigid bodies are elastic we can
More informationPurpose of this Guide: To thoroughly prepare students for the exact types of problems that will be on Exam 3.
ES230 STRENGTH OF MTERILS Exam 3 Study Guide Exam 3: Wednesday, March 8 th in-class Updated 3/3/17 Purpose of this Guide: To thoroughly prepare students for the exact types of problems that will be on
More informationThe University of Melbourne Engineering Mechanics
The University of Melbourne 436-291 Engineering Mechanics Tutorial Four Poisson s Ratio and Axial Loading Part A (Introductory) 1. (Problem 9-22 from Hibbeler - Statics and Mechanics of Materials) A short
More informationME 323 Examination #2 April 11, 2018
ME 2 Eamination #2 April, 2 PROBLEM NO. 25 points ma. A thin-walled pressure vessel is fabricated b welding together two, open-ended stainless-steel vessels along a 6 weld line. The welded vessel has an
More information3.032 Problem Set 1 Fall 2007 Due: Start of Lecture,
3.032 Problem Set 1 Fall 2007 Due: Start of Lecture, 09.14.07 1. The I35 bridge in Minneapolis collapsed in Summer 2007. The failure apparently occurred at a pin in the gusset plate of the truss supporting
More informationEquilibrium & Elasticity
PHYS 101 Previous Exam Problems CHAPTER 12 Equilibrium & Elasticity Static equilibrium Elasticity 1. A uniform steel bar of length 3.0 m and weight 20 N rests on two supports (A and B) at its ends. A block
More informationEDEXCEL NATIONAL CERTIFICATE/DIPLOMA MECHANICAL PRINCIPLES AND APPLICATIONS NQF LEVEL 3 OUTCOME 1 - LOADING SYSTEMS TUTORIAL 3 LOADED COMPONENTS
EDEXCEL NATIONAL CERTIICATE/DIPLOMA MECHANICAL PRINCIPLES AND APPLICATIONS NQ LEVEL 3 OUTCOME 1 - LOADING SYSTEMS TUTORIAL 3 LOADED COMPONENTS 1. Be able to determine the effects of loading in static engineering
More informationmportant nstructions to examiners: ) The answers should be examined by key words and not as word-to-word as given in the model answer scheme. ) The model answer and the answer written by candidate may
More informationand 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
More informationCHAPTER 3 THE EFFECTS OF FORCES ON MATERIALS
CHAPTER THE EFFECTS OF FORCES ON MATERIALS EXERCISE 1, Page 50 1. A rectangular bar having a cross-sectional area of 80 mm has a tensile force of 0 kn applied to it. Determine the stress in the bar. Stress
More informationIntroduction to Engineering Materials ENGR2000. Dr. Coates
Introduction to Engineering Materials ENGR2 Chapter 6: Mechanical Properties of Metals Dr. Coates 6.2 Concepts of Stress and Strain tension compression shear torsion Tension Tests The specimen is deformed
More informationReg. No. : Question Paper Code : B.Arch. DEGREE EXAMINATION, APRIL/MAY Second Semester AR 6201 MECHANICS OF STRUCTURES I
WK 4 Reg. No. : Question Paper Code : 71387 B.Arch. DEGREE EXAMINATION, APRIL/MAY 2017. Second Semester AR 6201 MECHANICS OF STRUCTURES I (Regulations 2013) Time : Three hours Maximum : 100 marks Answer
More informationCHAPTER 12 STATIC EQUILIBRIUM AND ELASTICITY. Conditions for static equilibrium Center of gravity (weight) Examples of static equilibrium
CHAPTER 12 STATIC EQUILIBRIUM AND ELASTICITY As previously defined, an object is in equilibrium when it is at rest or moving with constant velocity, i.e., with no net force acting on it. The following
More information2/28/2006 Statics ( F.Robilliard) 1
2/28/2006 Statics (.Robilliard) 1 Extended Bodies: In our discussion so far, we have considered essentially only point masses, under the action of forces. We now broaden our considerations to extended
More informationEntrance exam Master Course
- 1 - Guidelines for completion of test: On each page, fill in your name and your application code Each question has four answers while only one answer is correct. o Marked correct answer means 4 points
More informationSolution: T, A1, A2, A3, L1, L2, L3, E1, E2, E3, P are known Five equations in five unknowns, F1, F2, F3, ua and va
ME 323 Examination # 1 February 18, 2016 Name (Print) (Last) (First) Instructor PROBLEM #1 (20 points) A structure is constructed from members 1, 2 and 3, with these members made up of the same material
More informationstorage tank, or the hull of a ship at rest, is subjected to fluid pressure distributed over its surface.
Hydrostatic Forces on Submerged Plane Surfaces Hydrostatic forces mean forces exerted by fluid at rest. - A plate exposed to a liquid, such as a gate valve in a dam, the wall of a liquid storage tank,
More informationN = Shear stress / Shear strain
UNIT - I 1. What is meant by factor of safety? [A/M-15] It is the ratio between ultimate stress to the working stress. Factor of safety = Ultimate stress Permissible stress 2. Define Resilience. [A/M-15]
More informationSemester: BE 3 rd Subject :Mechanics of Solids ( ) Year: Faculty: Mr. Rohan S. Kariya. Tutorial 1
Semester: BE 3 rd Subject :Mechanics of Solids (2130003) Year: 2018-19 Faculty: Mr. Rohan S. Kariya Class: MA Tutorial 1 1 Define force and explain different type of force system with figures. 2 Explain
More informationFIXED BEAMS IN BENDING
FIXED BEAMS IN BENDING INTRODUCTION Fixed or built-in beams are commonly used in building construction because they possess high rigidity in comparison to simply supported beams. When a simply supported
More informationMECHANICS 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
More informationDirect and Shear Stress
Direct and Shear Stress 1 Direct & Shear Stress When a body is pulled by a tensile force or crushed by a compressive force, the loading is said to be direct. Direct stresses are also found to arise when
More informationQUESTION BANK DEPARTMENT: CIVIL SEMESTER: III SUBJECT CODE: CE2201 SUBJECT NAME: MECHANICS OF SOLIDS UNIT 1- STRESS AND STRAIN PART A
DEPARTMENT: CIVIL SUBJECT CODE: CE2201 QUESTION BANK SEMESTER: III SUBJECT NAME: MECHANICS OF SOLIDS UNIT 1- STRESS AND STRAIN PART A (2 Marks) 1. Define longitudinal strain and lateral strain. 2. State
More information2014 MECHANICS OF MATERIALS
R10 SET - 1 II. Tech I Semester Regular Examinations, March 2014 MEHNIS OF MTERILS (ivil Engineering) Time: 3 hours Max. Marks: 75 nswer any FIVE Questions ll Questions carry Equal Marks ~~~~~~~~~~~~~~~~~~~~~~~~~
More informationMECE 3321 MECHANICS OF SOLIDS CHAPTER 3
MECE 3321 MECHANICS OF SOLIDS CHAPTER 3 Samantha Ramirez TENSION AND COMPRESSION TESTS Tension and compression tests are used primarily to determine the relationship between σ avg and ε avg in any material.
More informationSolid 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
More informationMECHANICS OF MATERIALS
Fifth SI Edition CHTER 1 MECHNICS OF MTERILS Ferdinand. Beer E. Russell Johnston, Jr. John T. DeWolf David F. Mazurek Introduction Concept of Stress Lecture Notes: J. Walt Oler Teas Tech University Contents
More informationPORTMORE COMMUNITY COLLEGE ASSOCIATE DEGREE IN ENGINEERING TECHNOLOGY
PORTMORE COMMUNITY COLLEGE ASSOCIATE DEGREE IN ENGINEERING TECHNOLOGY RESIT EXAMINATIONS SEMESTER 2 JUNE 2011 COURSE NAME: Mechanical Engineering Science CODE: GROUP: ADET 1 DATE: JUNE 28 TIME: DURATION:
More informationTorque. Physics 6A. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB
Physics 6A Torque is what causes angular acceleration (just like a force causes linear acceleration) Torque is what causes angular acceleration (just like a force causes linear acceleration) For a torque
More informationEDEXCEL NATIONAL CERTIFICATE/DIPLOMA FURTHER MECHANICAL PRINCIPLES AND APPLICATIONS UNIT 11 - NQF LEVEL 3 OUTCOME 1 - FRAMES AND BEAMS
EDEXCEL NATIONAL CERTIFICATE/DIPLOMA FURTHER MECHANICAL PRINCIPLES AND APPLICATIONS UNIT 11 - NQF LEVEL 3 OUTCOME 1 - FRAMES AND BEAMS TUTORIAL 2 - BEAMS CONTENT Be able to determine the forces acting
More informationSECOND ENGINEER REG. III/2 APPLIED MECHANICS
SECOND ENGINEER REG. III/2 APPLIED MECHANICS LIST OF TOPICS Static s Friction Kinematics Dynamics Machines Strength of Materials Hydrostatics Hydrodynamics A STATICS 1 Solves problems involving forces
More informationPURE BENDING. If a simply supported beam carries two point loads of 10 kn as shown in the following figure, pure bending occurs at segment BC.
BENDING STRESS The effect of a bending moment applied to a cross-section of a beam is to induce a state of stress across that section. These stresses are known as bending stresses and they act normally
More informationProblem " 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 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
More informationUnit 1. (a) tan α = (b) tan α = (c) tan α = (d) tan α =
Unit 1 1. The subjects Engineering Mechanics deals with (a) Static (b) kinematics (c) Kinetics (d) All of the above 2. If the resultant of two forces P and Q is acting at an angle α with P, then (a) tan
More informationPHYSICS. Unit 3 Written examination Trial Examination SOLUTIONS
PHYSICS Unit 3 Written examination 1 2012 Trial Examination SECTION A Core Motion in one and two dimensions Question 1 SOLUTIONS Answer: 120 N Figure 1 shows that at t = 5 sec, the cart is travelling with
More informationWhere, m = slope of line = constant c = Intercept on y axis = effort required to start the machine
(ISO/IEC - 700-005 Certified) Model Answer: Summer 07 Code: 70 Important Instructions to examiners: ) The answers should be examined by key words and not as word-to-word as given in the model answer scheme.
More informationIn your answer, you should use appropriate technical terms, spelled correctly [1]
1 (a) Define moment of a force. In your answer, you should use appropriate technical terms, spelled correctly.... [1] (b) State the two conditions that apply when an object is in equilibrium. 1.... 2....
More informationAdvanced Structural Analysis EGF Section Properties and Bending
Advanced Structural Analysis EGF316 3. Section Properties and Bending 3.1 Loads in beams When we analyse beams, we need to consider various types of loads acting on them, for example, axial forces, shear
More informationCHAPTER 4: BENDING OF BEAMS
(74) CHAPTER 4: BENDING OF BEAMS This chapter will be devoted to the analysis of prismatic members subjected to equal and opposite couples M and M' acting in the same longitudinal plane. Such members are
More information[8] Bending and Shear Loading of Beams
[8] Bending and Shear Loading of Beams Page 1 of 28 [8] Bending and Shear Loading of Beams [8.1] Bending of Beams (will not be covered in class) [8.2] Bending Strain and Stress [8.3] Shear in Straight
More information(1) Brass, an alloy of copper and zinc, consists of 70% by volume of copper and 30% by volume of zinc.
PhysicsAndMathsTutor.com 1 Q1. (a) Define the density of a material....... (1) (b) Brass, an alloy of copper and zinc, consists of 70% by volume of copper and 30% by volume of zinc. density of copper =
More informationDownloaded from Downloaded from / 1
PURWANCHAL UNIVERSITY III SEMESTER FINAL EXAMINATION-2002 LEVEL : B. E. (Civil) SUBJECT: BEG256CI, Strength of Material Full Marks: 80 TIME: 03:00 hrs Pass marks: 32 Candidates are required to give their
More informationKINGS COLLEGE OF ENGINEERING DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK. Subject code/name: ME2254/STRENGTH OF MATERIALS Year/Sem:II / IV
KINGS COLLEGE OF ENGINEERING DEPARTMENT OF MECHANICAL ENGINEERING QUESTION BANK Subject code/name: ME2254/STRENGTH OF MATERIALS Year/Sem:II / IV UNIT I STRESS, STRAIN DEFORMATION OF SOLIDS PART A (2 MARKS)
More informationIndividual ASSIGNMENT Assignment 4: Moving charges, magnetic fields, Forces and Torques. Solution
Individual ASSIGNMENT Assignment 4: Moving charges, magnetic fields, Forces and Torques This homework must be solved individually. Solution 1. A sphere of mass M and radius R is suspended from a pivot
More information18.Define the term modulus of resilience. May/June Define Principal Stress. 20. Define Hydrostatic Pressure.
CE6306 STREGNTH OF MATERIALS Question Bank Unit-I STRESS, STRAIN, DEFORMATION OF SOLIDS PART-A 1. Define Poison s Ratio May/June 2009 2. What is thermal stress? May/June 2009 3. Estimate the load carried
More informationSample Question Paper
Scheme I Sample Question Paper Program Name : Mechanical Engineering Program Group Program Code : AE/ME/PG/PT/FG Semester : Third Course Title : Strength of Materials Marks : 70 Time: 3 Hrs. Instructions:
More informationSports biomechanics explores the relationship between the body motion, internal forces and external forces to optimize the sport performance.
What is biomechanics? Biomechanics is the field of study that makes use of the laws of physics and engineering concepts to describe motion of body segments, and the internal and external forces, which
More informationSubject : Engineering Mechanics Subject Code : 1704 Page No: 1 / 6 ----------------------------- Important Instructions to examiners: 1) The answers should be examined by key words and not as word-to-word
More informationThere are three main types of structure - mass, framed and shells.
STRUCTURES There are three main types of structure - mass, framed and shells. Mass structures perform due to their own weight. An example would be a dam. Frame structures resist loads due to the arrangement
More informationMoments practice exam questions
Moments practice exam questions Name - 50 minutes 48 marks Page 1 of 13 Q1. A waiter holds a tray horizontally in one hand between fingers and thumb as shown in the diagram. P, Q and W are the three forces
More informationOnly for Reference Page 1 of 18
Only for Reference www.civilpddc2013.weebly.com Page 1 of 18 Seat No.: Enrolment No. GUJARAT TECHNOLOGICAL UNIVERSITY PDDC - SEMESTER II EXAMINATION WINTER 2013 Subject Code: X20603 Date: 26-12-2013 Subject
More informationENG1001 Engineering Design 1
ENG1001 Engineering Design 1 Structure & Loads Determine forces that act on structures causing it to deform, bend, and stretch Forces push/pull on objects Structures are loaded by: > Dead loads permanent
More informationTwinning Engineering Programmes (TEP) & Thammasat English Programme of Engineering (TEPE) Faculty of Engineering, Thammasat University
" Twinning Engineering Programmes (TEP) & Thammasat English Programme of Engineering (TEPE) Faculty of Engineering, Thammasat University Undergraduate Examination 2 nd Semester of 2019 (Mid-term) CE221:
More informationFramed Structures PLANE FRAMES. Objectives:
Framed Structures 2 Objectives: ifferentiate between perfect, imperfect and redundant frames. To compute the member forces in a frame by graphical method. To compute the forces in a truss by method of
More informationStatic Equilibrium; Elasticity & Fracture
Static Equilibrium; Elasticity & Fracture The Conditions for Equilibrium Statics is concerned with the calculation of the forces acting on and within structures that are in equilibrium. An object with
More information(1) (allow e.c.f. for work done in (ii)) = 133 W (1) (allow e.c.f. for incorrect time conversion) 6
1. (a) F cos 20 = 300 gives F = 319 N 1 (i) work done = force distance moved in direction of force F is not in the direction of motion work done = force distance = 300 8000 = 2.4 10 6 J work done (iii)
More informationTUTORIAL SHEET 1. magnitude of P and the values of ø and θ. Ans: ø =74 0 and θ= 53 0
TUTORIAL SHEET 1 1. The rectangular platform is hinged at A and B and supported by a cable which passes over a frictionless hook at E. Knowing that the tension in the cable is 1349N, determine the moment
More informationEMA 3702 Mechanics & Materials Science (Mechanics of Materials) Chapter 2 Stress & Strain - Axial Loading
MA 3702 Mechanics & Materials Science (Mechanics of Materials) Chapter 2 Stress & Strain - Axial Loading MA 3702 Mechanics & Materials Science Zhe Cheng (2018) 2 Stress & Strain - Axial Loading Statics
More informationINTRODUCTION TO STRAIN
SIMPLE STRAIN INTRODUCTION TO STRAIN In general terms, Strain is a geometric quantity that measures the deformation of a body. There are two types of strain: normal strain: characterizes dimensional changes,
More informationChapter 2 Statics of Particles. Resultant of Two Forces 8/28/2014. The effects of forces on particles:
Chapter 2 Statics of Particles The effects of forces on particles: - replacing multiple forces acting on a particle with a single equivalent or resultant force, - relations between forces acting on a particle
More information