Chapter 1: Stress - Mechanics of Materials PDF

Chapter 1. Stress Outline 1.1 Introduction 1.2 Equilibrium of a Deformable Body 1.3 Stress 1.4 Average Normal Stress in an Axially Loaded Bar 1.5 Average Shear Stress 1.6 Allowable Shear Stress 1.7 Limit State Design Chapter Description Expected Outcomes - Identify and explain the concepts of stres...

Chapter 1. Stress Outline 1.1 Introduction 1.2 Equilibrium of a Deformable Body 1.3 Stress 1.4 Average Normal Stress in an Axially Loaded Bar 1.5 Average Shear Stress 1.6 Allowable Shear Stress 1.7 Limit State Design Chapter Description Expected Outcomes - Identify and explain the concepts of stresses. - Apply the concepts of stresses in calculating the normal and shear stress of body. - Analyse the average stress acting over the bar’s cross-sectional area. - Describe the single and double shear stresses thus, calculate the average shear stress. - Calculate the stress on oblique plane of body under loading conditions. - Calculate the allowable stress by applying the safety factors. 1.1 Introduction - Mechanics of materials is a study of the relationship between the external loads and internal loads within the body. - This subject also involves the deformations and stability of a body when subjected to external forces. - When a body subjected to an external load is sectioned, there is a distribution of force acting over the sectioned area which holds each segment of the body in equilibrium - The intensity of this internal force at a point in the body is referred to as stress - Stress is defined as force per unit area that the force acts upon 1.2 Equilibrium of a Deformable Body Equilibrium of a body requires a balance of forces and a balance of moments For a body with x, y, z coordinate system with origin O Best way to calculate these forces is to draw the free-body diagram (FBD). Internal Resultant Loadings - Internal resultants force – To hold the body together when the body subjected to external loads - Objective of FBD is to determine the resultant force and moment acting within a body. - In general, there are 4 different types of resultant loadings: a) Normal force, N b) Shear force, V c) Torsional moment or torque, T d) Bending moment, M Types of resultant loadings Free Body Diagram of the Internal Resultant Loadings Procedure for analysis STEP 1 – Support Reactions Decide which segment is to be considered Draw FBD Apply EOE STEP 2 – Draw FBD Draw FBD of the ‘cut’ segments, apply EOE and indicate the unknown resultants N, V, M and T STEP 3 – Equations of equilibrium Problem 1.6 Solution 1.3 Stress - Consider, a body subjected to external/internal force, F, acting on a cross sectional area A Stress can simply be defined as force divided by the cross section area. Unit: Pascal, Pa @ N/m2 Concept of stress Consider ΔA in figure below Small finite force, ΔF acts on ΔA Normal stress Intensity of force, or force per unit area, acting normal to ΔA Symbol used for normal stress, is σ (sigma) Tensile stress: normal force “pulls” or “stretches” the area element ΔA Compressive stress: normal force “pushes” or “compresses” area element ΔA Shear stress Intensity of force, or force per unit area, acting tangent to ΔA Symbol used for normal stress is τ (tau) 1.4 Average Normal Stress in An Axially Loaded Bar Examples of axially loaded bar Usually long and slender structural members Truss members, hangers, bolts Prismatic means all the cross sections are the same Assumptions 1. Uniform deformation: Bar remains straight before and after load is applied, and cross section remains flat or plane during deformation 2. In order for uniform deformation, force P be applied along centroidal axis of cross section Average normal stress distribution σ = average normal stress at any point on cross sectional area P = internal resultant normal force A = x-sectional area of the bar Equilibrium Consider vertical equilibrium of the element When the bar is stretched by the force P, the resulting stresses are tensile stresses If the force P cause the bar to be compressed, we obtain compressive stresses Sign convention for normal stresses is: (+) for tensile stresses and (-) for compressive stresses Because the normal stress σ is obtained by dividing the axial force by the cross– sectional area, it has units of force per unit of area. In SI units: Force is expressed in newtons (N) and area in square meters (m2). A N/m2 is a pascals (Pa). Maximum average normal stress - For problems where internal force P and x-sectional A were constant along the longitudinal axis of the bar, normal stress σ = P/A is also constant - If the bar is subjected to several external loads along its axis, change in x-sectional area may occur - Thus, it is important to find the maximum average normal stress To determine that, we need to find the location where ratio P/A is a maximum Maximum average normal stress - Draw an axial or normal force diagram (plot of P vs. its position x along bar’s length) - Sign convention: – P is positive (+) if it causes tension in the member – P is negative (−) if it causes compression - Identify the maximum average normal stress from the plot Procedure for Analysis Average normal stress Use equation of σ = P/A for x-sectional area of a member when section subjected to internal resultant force P Procedure for Analysis: Axially loaded members Internal Loading: - Section member perpendicular to its longitudinal axis at pt where normal stress is to be determined Draw free-body diagram - Use equation of force equilibrium to obtain internal axial force P at the section Average Normal Stress: - Determine member’s x-sectional area at the section - Compute average normal stress σ = P/A Example 1.6 Example 1.5 Solution 1.5 Average Shear Stress Shear stress is the stress component that act in the plane of the sectioned area. Consider a force F acting to the bar For rigid supports, and F is large enough, bar will deform and fail along the planes identified by AB and CD Free-body diagram indicates that shear force, V= F/2 be applied at both sections to ensure equilibrium Average shear stress over each section is: τavg = average shear stress at section, assumed to be same at each pt on the section V= internal resultant shear force at section determined from equations of equilibrium A = area of section Case discussed above is example of simple or direct shear Caused by the direct action of applied load F Occurs in various types of simple connections, e.g., bolts, pins, welded material Single shear - The cross-sectional area of bolt and bonding surface between the two members are subjected to one connection (single shear force). i.e, for equilibrium, V = F - The joint shown below illustrate typical examples of a single shear. Double shear - The cross-sectional area of bolt and bonding surface between two members subjected to two connections (double shear force) i.e., for equilibrium, V = F/2 - The joints shown below are examples of double shear connections. Multiple shear - The cross-sectional area of bolt and bonding surface between two members subjected to multiple shear with n, connections i.e., for equilibrium, V = F/n - The joints shown below are examples of multiple shear connections. Procedure for analysis Resultant Internal shear force 1. Cut the body where the average shear stress, τavg is to be calculated 2. Draw free-body diagram 3. Use equation of equilibrium (EoE) to compute the internal shear force V Average shear stress 1. Calculate the cross-sectional area A 2. Compute average shear stress using τavg = V/A Example 1 A lap joint is fastened together using six bolts as depicted in the figure below. Each of the plate thickness, t = 10 mm. If the diameter of the bolts is 20 mm. Determine the average shear stress on the bolts. Solution Example 2 Depth and thickness = 40 mm, determine average normal stress and average shear stress acting along (a) section planes a-a, and (b) section plane b-b. Solution Part (a) Internal loading Based on free-body diagram, Resultant loading of axial force, P = 800 N Average stress Average normal stress, σ Part (b) Internal loading Allowable Stress - consideration when designing a structural elements or mechanical components, is to design a member will not fail during useful operation - In order to achieve this, the stress in the member must be set at a level where the member will be safe - Select an allowable load that is less than the failure load of the member - The common approach used is known as factor of safety (F.S) - F.S. can also be expressed in terms of stress acting on the member as given below: Note: To avoid potential failure of the member, F.S. must be chosen to be greater than 1. However, the actual value will depend on the types of material used and the potential application of the member. Design of Simple Connections - To design a simple connection, one of the key parameter is the cross-sectional area over which critical stress is acting. - To obtain the area of member subjected to a normal force, - To obtain the area of member subjected to a shear force, Step for analysis First consider the section where the critical stress is acting. When using average normal stress and shear stress equations Internal Loading 1. Cut the member at the location where the critical stress is acting 2. Draw free-body diagram (FBD) 3. Use equations of equilibrium (EoE) to obtain internal resultant force Required Area From the given allowable stress, compute the required area to withstand the load using DESIGN OF SIMPLE CONNECTIONS Cross-sectional area of a tension member Condition: The force has a line of action that passes through the centroid of the cross section. Cross-sectional area of a connecter subjected to shear Assumption: If bolt is loose or clamping force of bolt is unknown, assume frictional force between plates to be negligible. Example The two members pinned together at B. If the pins have an allowable shear stress of τallow = 90 MPa, and allowable tensile stress of rod CB is (σt)allow = 115 MPa Determine to nearest mm the smallest diameter of pins A and B and the diameter of rod CB necessary to support the load. Solution Draw free-body diagram: Diameter of pins: Diameter of pins: Choose a size larger to nearest millimeter. Diameter of rod: Problem 1-19 Solution: Addition question: Determine average normal stress and average shear stress at D, area D is 0.0004 m2 Soltution: Problem 1-37 Solution Problem 1-47 Solution