CONCURRENT FORCES & FORCE DIAGRAMS  A Concurrent Force System, is a force system where all forces start from the same point. 100N 200N  TERMS:  Resultant,

Slides:



Advertisements
Similar presentations
BENDING MOMENTS AND SHEARING FORCES IN BEAMS
Advertisements

BTECH Mechanical principles and applications
Aim: How can we describe resultant force (net force)?
Axial Members WORKSHEET 11 to answer just click on the button or image related to the answer.
CTC / MTC 222 Strength of Materials Chapter 12 Pressure Vessels.
Analysis of Beams in Bending ( )
Course Title: Strength of Materials (CVE 202)
Chapter 6 Bending.
Tutorial 4 MECH 101 Liang Tengfei Office phone : Mobile : Office hour : 14:00-15:00 Fri 1.
Read Chapter 1 Basic Elasticity - Equilibrium Equations
A 10-m long steel wire (cross – section 1cm 2. Young's modulus 2 x N/m 2 ) is subjected to a load of N. How much will the wire stretch under.
CTC / MTC 222 Strength of Materials
ENGR 220 Section 6.1~6.2 BENDING.
Professor Joe Greene CSU, CHICO
ENGR 225 Section 2.1. Review ~ Force per Area Normal Stress –The average normal stress can be determined from σ = P / A, where P is the internal axial.
ECIV 320 Structural Analysis I Internal Loadings in Structural Members Sections 4.1 – 4.5 Study all examples.
CM 197 Mechanics of Materials Chap 18: Combined Stresses
Copyright © 2011 Pearson Education South Asia Pte Ltd
Copyright © 2011 Pearson Education South Asia Pte Ltd
BFC (Mechanics of Materials) Chapter 2: Shear Force and Bending Moment
BEAMS SHEAR AND MOMENT.
Beams Beams: Comparison with trusses, plates t
Beam Analysis Civil Engineering and Architecture
Bending Shear and Moment Diagram, Graphical method to construct shear
Shear Forces & Bending Moments Shear & Moment Diagrams
MECHANICS OF MATERIALS 7th Edition
Shear Forces and Bending Moments in Beams
Aim: How do we add force vectors? What is an equilibrant vector? Do Now: Sam and Joey are fighting over a toy. Sam pulls with a force of 30N to the left.
7.2 Shear and Moment Equations and Diagrams
SHEAR AND BENDING MOMENT DIAGRAMS IN HORIZONTAL BEAMS WITH
Objectives Students will be able to label a stress-strain diagram correctly indicating. Ultimate stress, yield stress and proportional limit. Students.
 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : UNIAXIAL LOAD 1 PROBLEM-1 1 m P A composite A-36 steel bar shown in the figure has 2 segments,
Axial Members AXIAL MEMBERS, which support load only along their primary axis, are the most basic of structural members. Equilibrium requires that forces.
Chapter 1: Stress Review important principles of statics
Buckling of Slender Columns ( )
ENT 153 TUTORIAL 1.
Eng. Tamer Eshtawi First Semester
 2005 Pearson Education South Asia Pte Ltd TUTORIAL-1 : STRESS & STRAIN 1 PROBLEM-1 The hanger assembly is used to support a distributed loading of w=16kN/m.
CTC / MTC 322 Strength of Materials
Principle of Engineering ENG2301 F Mechanics Section F Textbook: F A Foundation Course in Statics and Dynamics F Addison Wesley Longman 1997.
Tension & Compression Labs SMTL 246. These are the results from one of the tension labs.
A simply supported beam of span 8 m carries two concentrated loads of 32 kN and 48 kN at 3m and 6 m from left support. Calculate the deflection at the.
Dean Hackett Structures Week 2. Dean Hackett In previous sessions… Brief review of previous learning: –Types of motion –Classes of lever –Turning moments.
6- Calculation of shear stress at composite interface: A)Under service load: Strain and stress distributions across composite beam cross- section, under.
Unit-5. Torsion in Shafts and Buckling of Axially Loaded Columns Lecture Number-3 Mr. M.A.Mohite Mechanical Engineering S.I.T., Lonavala.
CONTINUATION OF COMPONENTS OF FORCES Realize in these problems that a right triangle will represent a FORCE and the COMPONENTS of the force, when the.
Triaxial State of Stress at any Critical Point in a Loaded Body
Thermal Stress Objective: Lecture # 8
Axial Force Definition: Force which is parallel to the longitudinal axis of the member.
Tutorial 7_Review MECH 101 Liang Tengfei Office phone : Mobile : Office hour : 14:00-15:00 Fri 1.
ES2501: Statics/Unit 23-1: Internal Forces in Beams: More Examples
Sample Problem 4.2 SOLUTION:
Shearing Stresses in Beams and Thin-Walled Members
Vectors.
Chapter 1 Stress and Strain.
Ch. 2: Fundamental of Structure
Structure I Course Code: ARCH 208 Dr. Aeid A. Abdulrazeg
Sample Problem 4.2 SOLUTION:
Problem-1 A two member frame is supported by the two pin supports at A and D as shown. The beam AB is subjected to a load of 4 kN at its free end. Draw.
Shearing Stresses in Beams and Thin-Walled Members
Chapter 5 Torsion.
Chapter 6 Bending.
4.7 STRESS CONCENTRATIONS
ANALYSIS OF STRUCTURES
Examples on Equilibrium
What is a beam? Structural member that supports a load
Copyright ©2014 Pearson Education, All Rights Reserved
Physics Chapter 6 Equilibrium.
Why to Provide Redundant Members?
Tutorial.
Presentation transcript:

CONCURRENT FORCES & FORCE DIAGRAMS  A Concurrent Force System, is a force system where all forces start from the same point. 100N 200N  TERMS:  Resultant, the single force which can replace a force system.  Equilibrant, the single force which can bring a force system in to equilibrium.  To achieve equilibrium, all horizontal, vertical and moments must be balanced.  We can determine the value of the resultant and equilibrant through graphical techniques.

CONCURRENT FORCES & FORCE DIAGRAMS Lets consider the previous force system. 100N 200N 1.Select a scale to draw the forces against. In this case, lets say 5mm = 10N. 2.Draw the 200N force to scale, i.e. 100mm. 3.Draw the 100N force from the end of this vertically upwards. i.e. 50mm. 4. Join the ends and measure this line. 100mm 50mm Resultant Line length = Approx. 110mm So resultant = (110/5) x 10N = Approx. 220N

CONCURRENT FORCES & FORCE DIAGRAMS The Equlibrant will always be in the opposite direction, but will always have the same value. This technique works for force systems with more than 2 forces, as an example. F1 F2 F3 F4 Force System F1 F2 F3 F4 R

CONCURRENT FORCES & FORCE DIAGRAMS Pupil problems: N 30N 1)2) 50N 45 65N 3) 150N 50N 200N 75N

STRESS & STRAIN STRESS = Load/Cross Sectional Area (Nmm-2) Strain = Change in Length/ Original Length (no units) Example 1: A structural member is subjected to a compressive force of 20000N. If the diameter of the member is 250mm, calculate the stress within the member. If the load reduces the member length by 0.5mm and the original length was 2m, calculate the strain.

Solutions Strain Strain = Change in Length/ Original Length = 0.5/2000 =