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2/24/2021 CMCE 1215 Strength of Materials Spring 2021 Bring to class: Tensile sample Dial gage PROPERTIES OF MATERIALS I PROF. I VAN L. GUZMAN, PHD, MBA, PE NEW YORK CITY COLLEGE OF TECHNOLOGY AGENDA Stress vs Strain Diagram I.     Define Proportional Limit Define Yield Point Define Ultimate Strength Define Rupture Strength II. Define Ductility III. Example IV. Project I 1 2/24/2021 In the area of strength of materials, problems are primarily of two types: analysis and design. • Problems of analysis could involve finding the greatest load that may be applied to a given body without exceeding specified limiting values of stress and strain. • Problems of design involve determining the required size and shape of a member to support given loads without exceeding specified limiting stress and/or strain. The mechanical properties (of a given material) are important to our study since they are the properties that affect the limiting values of stress and strain. Limbrunner I. Tensile Test The simplest way of establishing relevant mechanical properties of a vast array of materials is the Tensile Test. In which consecutive tensile loads are applied and changes in length (deformation) recorded. • The load and deformation readings are then expressed as stress and strain, and then plotted. 2 2/24/2021 I. Tensile Test (cont.) The simplest way of establishing relevant the mechanical properties of a vast array of materials is the Tensile Test. In which consecutive tensile loads are applied and changes in length (deformation) recorded. • The load and deformation readings are then expressed as stress and strain and plotted. P σ δ ε What is a displacement gage and gage length? • A displacement gage (dial gage) is an instrument that is attached to the specimen to measure deformation (displacement) during loading. • The Gage length is the distance between the points of contact of the gage on the specimen. • Dial gages can be mechanical or digital. 3 2/24/2021 I. Tensile Test (cont.) From the plot the following mechanical properties can be evaluated. ■ “A” Proportional Limit – The Proportional limit is the limit of linear elasticity ■ “B” Elastic Limit – The elastic limit maybe defined as the greatest stress a material is capable of developing without a permanent elongation. ■ “C” Yield Point – The point at which Strain increases at a faster rate and continues with little or no increase in applied load or stress. The stress at this point is called Yield Stress (Fy). ■ “D” Lower Yield Point – Abrupt decrease in stress. Typical of low carbon steel (not all steels). ■ “E” Ultimate Strength – The maximum stress a material is capable of developing. (Fu) I. ■ “F”Rupture Strength – After the stress passes point E, the steel begins to decrease in diameter and increase in length. This is the point at which the specimen breaks. ■ “G” True Stress at failure – Rupture stress calculated using final area. ■ Modulus of Elasticity – Slope of the linear range of the curve (anywhere between “0” and “A” in the diagram) Tensile Test (cont.) Yield Point or Yield Stress (alternate): If a material does not a exhibit a welldefined yield point during tensile tests, it can be approximated by using the 0.2% (0.002 in/in) offset method. 4 2/24/2021 II. Ductility  The property of a material enabling it to undergo considerable plastic deformation under tensile load before rupture.  Ductility is defined by the percent elongation of the specimen gage length in the direction of loading and by the percent reduction in area. – High percent elongation indicates high ductility material (>5%) Example #1 GIVEN: From a laboratory tensile test the following data was recorded: ■ Original Specimen Gage Length: 2″ ■ Final Specimen Gage Length: 2 – 7/16″ ■ Original Diameter: 0.493″ ■ Final Diameter: 0.325″ ■ Yield Load: 9,300 # ■ Ultimate Load: 12,000 # ■ Rupture Load: 11,100 # ■ Gage Factor: 1 division = .0001“ REQUIRED: 1. Plot the Stress Strain Graph 2. Determine the Critical Values: a. Modulus of Elasticity “E” b. Stress @ the Proportional Limit “FPL” c. Stress @ the Yield Load : “Fy” d. Stress @ the Rupture Load: “FRUPT“ e. True stress at failure f. Ductility– Elongation Delongation 5 2/24/2021 What are gage divisions? Example #1 Load (lbs) Gage (div) 0 0 1000 8 2000 20 3000 31 4000 43 5000 56 6000 70 7000 89 8000 125 8500 150 Elongation (in) Stress (ksi) Strain What are gage divisions? What is the gage factor? What is the displacement reading for this case? 6 2/24/2021 PROJECT I *Blackboard submission due 3/11/21 at beginning of class.  References:  Applied Statics and Strength of Materials, Limbrunner and D’Allaird  Special  Thanks to: Prof. Anthony Cioffi www.faithec.org 7 3/3/2021 CMCE 1215 Strength of Materials Spring 2021 Bring to class: PROPERTIES OF MATERIALS II PROF. I VAN L. GUZMAN, PHD, MBA, PE NEW YORK CITY COLLEGE OF TECHNOLOGY 3/1/21 http://khc.qcc.cuny.edu/events-2/ 1 3/3/2021 AGENDA Lecture: I. II. I. Poisson’s Ratio II. Thermal Effects III. Stresses on Inclined Planes Assignment I. Poisson’s Ratio – – The ratio of lateral strain to axial strain. A measure of the degree to which a material expands outwards when squeezed, or equivalently contracts when stretched. 𝝁= 𝜺𝒕𝒓𝒂𝒏𝒔𝒗𝒆𝒓𝒔𝒆 𝜺𝒂𝒙𝒊𝒖𝒂𝒍 2 3/3/2021 I. Poisson’s Ratio – Typical Values Example 1: Given: 2” diameter rod Axial Tensile Load = 40kips ɛlongitudinal = 0.0012 (strain) ɛtransverse = 0.0004 (strain) Required: 1. Poisson’s Ratio (μ) 2. Modulus of Elasticity (E) 3 3/3/2021 II. • • • • • Thermal Effects Most substances expand when heated and contract when cooled. The change in length / area or volume (due to contracting/expanding) is directly related to temperature change. However, not all materials contract/expand at the same rate; the amount of contraction/expansion varies depending on the material. The linear coefficient of thermal expansion (α), is a proportionality constant that determines the rate of change in length in different materials when they are heated and cooled. The Total Change in Length: δ = αL x ∆ T (inches or mm) Definitions: L = Original Length α = The linear coefficient of thermal o o Expansion (1/ F or 1/ C) ∆T = The change in Temperature o o units: F or C Fire Damage II. Thermal Effects (cont.) • Thermal Stresses: The stress developed in a restrained member due to a change in temperature. • By equating the two change in length formulas we have developed to date we can solve for: 1 𝑃𝐿 𝛿 = 𝐴𝐸 = 𝜎 𝐿 𝐸 and 2 δ = α L (∆ T)  σ = α E (∆ T) 4 3/3/2021 Example #2 a) Compute the thermal stresses in the following restrained steel beam if there was an overnight change in temperature o o from 80 F to 20 F. b) If the proportional limit of the steel is 34,000 psi will there be any permanent damage to the section? c) Is the steel member in tension of compression? Example #3 An AISI 1040 steel fence wire 0.148 inches in diameter is stretched between rigid end posts with a tension of 300 lb when the temperature is 90°F. The proportional limit of the wire is 40,000 psi. a) Calculate the temperature drop that could occur without causing a permanent set in the wire. 5 3/3/2021 III. Stresses on Inclined Plane • When a member is subjected to an axial Force (tensile or compressive), a maximum stress (tensile or compressive) is developed on a plane perpendicular (normal) to the longitudinal axis of the member. • Additional (smaller) Tensile/Compressive and Shear stresses are also developed on planes inclined to the cross-section. Element subjected to Axial Load: • Two Types of Stress in all elements: • • Normal Stress: σ (Ten. or Comp.) Shear Stress: τ (caused by Comp. or Ten) 6 3/3/2021 Element subjected to Axial Load: • Two Types of Stress in all elements: • • Normal Stress: σ (Ten. or Comp.) Shear Stress: τ (caused by Comp. or Ten) This expression becomes a maximum when cos2 = 1, i.e. when  = 0o This expression becomes a maximum when sin2 = 1, i.e. when  =45o For the case of Pure Shear Stress – For the case of Pure Shear Stress Based on Equilibrium: τ1 = τ2 7 3/3/2021 Element Subjected to Pure Shear Stress: • Tension & Compression Stresses are created Normal Stress – Diagonal Surface σn = τ sin 2 Shear Stress – Diagonal Surface τ’ = τ cos 2 Definitions τ = Shear Stress τ’ = Shear Stress acting on a diagonal surface τ’ = τ cos 2 σn = τ sin 2 Example #4 A Shear stress (pure shear) of 8,000 psi exists in a member. Compute the shear and Normal Stress developed on a diagonal surface whose normal are oriented at angles 15˚, 30˚ 45˚ with the longitudinal axis. 8 3/3/2021 ASSIGNMENT Complete the following problems from the textbook (7th Edition): Assignment #3 = 11.1,11.13,11.15,11.16,11.60 • *Due 3/11/21 at the beginning of class Could not find Error 11.60 normal stress should be 1500 psi 9 3/3/2021  References:  Applied Statics and Strength of Materials, Limbrunner and D’Allaird  Special  Thanks to: Prof. Anthony Cioffi www.faithec.org 10 3/9/2021 CMCE 1215 Strength of Materials Spring 2021 STATE OF STRESSES INSIDE OF AN ELEMENT PROF. I VAN L. GUZMAN, PHD, MBA, PE NEW YORK CITY COLLEGE OF TECHNOLOGY AGENDA I. Attendance II. Schedule III. Element Subject to Axial Load Mohr’s Circle IV. I. II. V. Commons Cases Peculiarities of the Mohr’s Circle Practice Problems 1 3/9/2021 Element subjected to Axial Load: • Two Types of Stress in all elements: • • Normal Stress: σ (Ten. or Comp.) Shear Stress: τ (caused by Comp. or Ten) 2. Mohr Circle: • A German engineer, Otto Mohr developed a graphical method to assist in visualizing the state of stresses (Normal and Shear) inside of a body. • It is widely known as the Mohr’s Circle. The Mohr’s Circle is a graphical representation of the transformation equations for plane stresses. 2 3/9/2021 2. Mohr Circle (cont.): Common Cases for analysis: A) Stresses developed inside a member with uniaxial loading. (CMCE 1215) B) Stresses developed inside a member with biaxial loading (normal Stresses). (CMCE 2456, 2315, 2416) C) Stresses developed inside a member with biaxial loading (normal and shear stresses). (CMCE 2456) 2. Mohr Circle (cont.): A) Stresses developed inside a member with uniaxial loading. (CMCE 1215) • In this application we will determine the normal and shear stresses acting on an inclined plane of an axially loaded member using the Mohr’s circle. • Our goal is to find the normal (σ) and shear stresses (τ) on a plane that is inclined at an angle () 1. First step is to represent the external loads on an element inside of the member. 3 3/9/2021 2. Mohr Circle (cont.): A) Stresses developed inside a member with uniaxial loading. (CMCE 1215) How to draw a More Circle: 1. First step is to represent the external loads, on an element inside of the member. (from previous slide) 2. Establish a rectangular coordinate system in which normal stresses (σ) are plotted in the abscissa (horizontal axis) and the shear stresses (τ) are plotted in the ordinate (vertical axis). 3. Establish a scale that is the same for both axis. 4. Locate the stress combinations (σ, τ) found in the elements individual planes (i.e. 2 combinations for 2-D problem). 5. Plot the stress combinations (σ, τ). These points are part of the outer rim of the circle. 6. Draw a line between these two points. The center of that line will be the center of the circle. Sign Convention: Normal stress (σ): (+) Tension. (-) Compression Shear stress (τ): (+) CW, (-) CCW 2. Mohr Circle (cont.): A) Stresses developed inside a member with uniaxial loading. (CMCE 1215) (cont.) Procedure (cont.): 7. Using the center of the circle and one of the points on the outer rim, use a compass to draw the circle. You have completed the Mohr’s Circle!!!! 8. To find stresses acting at an angle (). From a known point on the circle (known stress combination), rotate graphically twice the angle (2 x ) in the same direction and draw a line that passes through the center. 9. The location were the line intersects the circle gives us the stress combination at the plane in question. Sign Convention: Normal stress (σ): (+) Tension. (-) Compression Shear stress (τ): (+) CW, (-) CCW 4 3/9/2021 2. Mohr Circle (cont.): A) Stresses developed inside a member with uniaxial loading. (CMCE 1215) Other Peculiarities of the Mohr’s Circle: • If on a given plane the shear stresses are zero (σ, 0), these planes are called Principle planes. • They always exist in any element. • The stress combinations for principle planes, always plots on the x axis (σ, 0) • The top of the circle represents the maximum shear stress found inside of an element. • If the uniaxial normal stress is in compression (negative) , the circle plots partially or completely to the left of the yaxis. Sign Convention: Normal stress (σ): (+) Tension. (-) Compression Shear stress (τ): (+) CW, (-) CCW Example 1 A 10-in.-long metal block, 4 in. by 4 in. in cross section, is subjected to a tensile load of 32,000 lb. Using Mohr’s circle, (a) determine the normal (tensile or compressive) stress and shear stress on a plane that is inclined 25° clockwise from the X–X axis, as shown in the figure, and (b) determine the magnitude of the maximum shear stress. 5 3/9/2021 PRACTICE PROBLEMS • Complete the following problems from the textbook (7th Ed): 17.42 & 17.43  References:  Applied Statics and Strength of Materials, Limbrunner and D’Allaird www.faithec.org 6 CMCE 1215 Strength of Materials Spring 2021 Stress Bring to class: W beam 2” x 4” Steel Manual Book Bolted Connections Prof. Ivan L. Guzman, PhD, MBA, PE New York City College of Technology 2 Agenda I. Introduction: Properties of Materials II. Force & Stress A. B. C. D. Types of Forces Unit Stress Normal Stress Shear Stress III. Stress in Bolted Connections and Plates A. B. C. Design / Failure Considerations Allowable Stresses Analysis A. B. D. Single Shear Double Shear Punching Shear 1.1 3 PROPERTIES OF MATERIALS I. INTRODUCTION ► Strengths of Materials is the STUDY of the relationships between external forces which act on Elastic Bodies and the Internal Stresses and Strains caused by these forces. • External Forces produce Internal Stresses & Strains • Deformation & Dimensional changes occur when an object is loaded. 4 II. FORCE & STRESS A. Types of Forces • • • • • Compression Tension Torsion Bending Shear Shear Units – lbs (pds, #), Kips, Tons, Newtons (N), Kilonewtons (kN)…. Shear 1.2 5 B. Unit Stress ►Normal (Direct) Stress – Stress acting on a plane perpendicular (normal) to the line of action of the applied force. ►Axial Stress – acting along the axis of the member. σ= P/A σ = average normal stress P = external applied load or force A = cross-sectional area perpendicular to the force • Units of Stress: – – – – – psi = lb/in^2, psf = lb/ft^2 ksi = kips/in^2, ksf = kips/ft^2 Pa = N/m^2 (Pascal) kPa = kN/m^2 (kilo Pascal) MPa = MN/m^2, N/mm^2 (Mega Pascal) C. Normal Stress – Tension and Compression ■ Tension “Pull” Tensile forces cause elongation (stretch) of a member. ■ Compression “Push” Compressive forces cause shortening of a member. Produces Tensional Stress Produces Compressional Stress 1.3 7 Example 1: Wooden Post Compute the compressive (normal) stress for a 2” x 4” nominal wooden post that is subjected to an axial compressive load of 25 kips. Properties of Structural Timber “Nominal” size refers to the rough-cut green wood. After drying and planning the timber member will reduce in size, and is then called the “Dressed” size. ** See appendix E of your textbook for additional typical sizes 1.4 9 Example 2: Structural Steel Compute the tensile (normal) stress for a W27 x114 structural steel member subjected to an axial tensile load of 600 kips. 10 Properties of Structural Steel Nomenclature: *For example, a W24 x162 has an approximate Depth (d) of 24 in and a weight per linear foot of 162 lbs/ft 1.5 Properties of Structural Steel (Contd.) *More shapes and sections on Appendix A, B, C & D of your textbook or in the Steel Manual. 12 • Other Typical Steel Sections: http://www.aboutcivil.org/steel-structure-types-tension-compression-trusses-shell.html 1.6 13 Example 3: A flat steel bar, 1/2 in. thick and 4 in. wide, is subjected to a 20-kip tensile load. Two 3/4 in diameter holes are located as shown. Required: Determine the average tensile stress at sections A–A and B–B. Neglect stress concentrations adjacent to each hole. D. Shear Stress ■ Stress that develops in a direction parallel to direction of loading. *Units: Same as axial stress σv = P/A σv = average shear stress(other symbols, P = external applied force A = cross-sectional area parallel to the force’s line of action (area where shear stress develops) τ, ss) 1.7 15 Example 4: Two wood blocks are glued together as shown. The dimensions of each wood block are 2 ft x 1.5 ft x 0.5 ft (l x w x h). If a 15 k load is applied as shown, what is the shear stress? 16 III. Stress in Bolted Connections & Plates A. Design/ Failure Considerations: a) Shear Failure in Bolts – CMCE 1215 b) Shear Failure in Plate – CMCE 2315 c) Bearing Failure in Bolts – CMCE 2315 d) Bearing Failure in the Plates – CMCE 2315 e) Tension Failure in Bolts – CMCE 2315 f) Tension Failure in the Plates – CMCE 2315 Bolted Connectio 1.8 17 B. Allowable Stresses: AISC Reference: J3.2 Allowable Stresses Allowable stresses are specified for: ■ Shear “σv” (AISC) ■ Bearing “σb” – CMCE 2315 ■ Tension “σt“ – CMCE 2315 C. Analysis: Shear on the Bolts: Bolts can be in either: P ►Single Shear – One Shear Plane ►Double Shear – Two Shear Planes 18 I. Single Shear ►General Equation: Pv = Ab σv Definitions: Pv : Allowable Shear Force (Load) (Single Bolt) Ab: Cross Sectional Area of the Bolt: AISC σv : Allowable Shear Stress: AISC II. Double Shear ►General Equation: Pv = 2(Ab σv) Note: σv depends on the Type of High Strength Bolt, Type of Connection and the Type of Hole (Standard, Oversized, Slotted) 1.9 EXAMPLE 5: Bolt Shear Two bolts are used to hold three plates together. The bolts are ¾” in diameter and a tensile load of 18,000 lb is applied. Required: Find the shear stress (σv) in each bolt. 20 D. Punching Shear σv= P/A • σv = average shear stress • P = external applied force or load • A = cross-sectional shear area perpendicular to the force ■ Units: Same as axial stress How would you compute area for the case presented??? 1.10 EXAMPLE 6: Punching Shear A punching operation is being planned in which a 34 in. diameter hole is to be punched in an aluminum plate 1.4in. thick. For this aluminum alloy, the ultimate shear strength is 27,000 psi. Required: Compute the force P that must be applied to the punch. Assume that the shear stress is uniformly distributed. 22 Assignment • Complete problems from the 7th Edition of the texbook: 9.3, 9.9, 9.10, 9.14, 9.17, 9.18, 9.20, & 9.27 • Due a week from today (2/18/21), at the beginning of class. 1.11 23 • References: – Applied Statics and Strength of Materials, Limbrunner and D’Allaird, 7th Edition • Special Thanks to: – Prof. Anthony Cioffi www.faithec.org 1.12 2/16/2021 CMCE 1215 Strength of Materials Spring 2021 Bring: Something that deforms elastical…

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