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Normal Stress (1.1-1.5) MAE 314 – Solid Mechanics Yun Jing Normal Stress 1 Pretest This is a structure which was designed to support a 30kN load, it consists of a boom AB and of a rod BC. The boom and rod are connected by a pin at B and are supported by pins and brackets at A and C. (1) Is there a reaction moment at A? why? (2) What is the reaction force in the vertical direction at A? (3) What is the internal 2 force in AB? (4) What is the internal force in BC? Statics Review Pins = no rxn moment Normal Stress 3 Statics Review • Solve for reactions at A & C: M C 0 Ax 0.6 m 30 kN 0.8 m Ax 40 kN Fx 0 Ax C x C x Ax 40 kN Fy 0 Ay C y 30 kN 0 Ay C y 30 kN • Ay and Cy can not be determined from these equations. Normal Stress 4 Statics Review • Consider a free-body diagram for the boom: M B 0 Ay 0.8 m Ay 0 substitute into the structure equilibrium equation Bx C y 30 kN • Results: A 40 kN Cx 40 kN C y 30 kN See section 1.2 in text for complete static analysis and review of method of joints. Normal Stress 5 Normal Stress 6 Introduction to Normal Stress Methods of statics allow us to determine forces and moments in a structure, but how do we determine if a load can be safely supported? Factors: material, size, etc. Need a new concept….Stress Normal Stress 7 Introduction to Normal Stress Stress = Force per unit area F A Normal Stress 8 Introduction to Normal Stress If stress varies over a cross-section, the resultant of the internal forces for an axially loaded member is normal to a section cut perpendicular to the axis. Thus, we can write the stress at a point as lim A0 F A We assume the force F is evenly distributed over the cross-section of the bar. In reality F = resultant force over the end of the bar. dA F A Normal Stress 9 Introduction to Normal Stress Sign convention 0 0 Tensile (member is in tension) Compressive (member is in compression) Units (force/area) English: lb/in2 = psi kip/in2 = ksi SI: Tensile N/m2 = Pa (Pascal) kN/m2 = kPa MPa, GPa, etc. Compressive Normal Stress 10 Definitions and Assumptions Homogenous: material is the same throughout the bar Cross-section: section perpendicular to longitudinal axis of bar A F’ F Prismatic: cross-section does not change along axis of bar Prismatic Non-Prismatic Normal Stress 11 Definitions and Assumptions Uniaxial bar: a bar with only one axis Normal Stress (σ): stress acting perpendicular to the cross-section. Deformation of the bar is uniform throughout. (Uniform Stress State) Stress is measured far from the point of application. Loads must act through the centroid of the cross-section. Normal Stress 12 Definitions and Assumptions The uniform stress state does not apply near the ends of the bar. Assume the distribution of normal stresses in an axially loaded member is uniform, except in the immediate vicinity of the points of application of the loads (Saint-Venant’s Principle). Normal Stress “Uniform” Stress 13 Definitions and Assumptions How do we know all loads must act through the centroid of the cross-section? Let us represent P, the resultant force, by a uniform stress over the cross-section (so that they are statically equivalent). Normal Stress 14 Definitions and Assumptions Moments due to σ: Mx ydA A M y xdA A M M Set M My y Mx = Mx x and MMy = y x 1 1 y ydA ydA PA AA Equations for the centroid 1 1 x xdA xdA PA AA Normal Stress 15 Example Problem Can the structure we used for our statics review safely support a 30 kN load? (Assume the entire structure is made of steel with a maximum allowable stress σall=165 MPa.) Cross-section 30 mm x 50 mm Normal Stress 16 Example Problem Two cylindrical rods are welded together and loaded as shown. Find the normal stress at the midsection of each rod. d1 50 mm d 2 30 mm Normal Stress 17 Shearing and Bearing Stress (1.6-1.8, 1.12) MAE 314 – Solid Mechanics Yun Jing Shearing and Bearing Stress 18 What is Shearing Stress? We learned about normal stress (σ), which acts perpendicular to the cross-section. Normal stress results in a volume change. Shear stress (τ) acts tangential to the surface of a material element. Shear stress results in a shape change. Shearing and Bearing Stress 19 Where Do Shearing Stresses Occur? Shearing stresses are commonly found in bolts, pins, and rivets. Bolt is in “single” shear Free Body Diagram of Bolt Force P results in shearing stress Force F results in bearing stress (will discuss later) Shearing and Bearing Stress 20 Shear Stress Defined We do not assume τ is uniform over the cross-section, because this is not the case. τ is the average shear stress. ave P F A A The maximum value of τ may be considerably greater than τave, which is important for design purposes. Shearing and Bearing Stress 21 Double Shear Bolt is in “double” shear Free Body Diagram of Bolt Free Body Diagram of Center of Bolt ave Shearing and Bearing Stress F P F 2 A A 2A 22 Bearing Stress Bearing stress is a normal stress, not a shearing stress. Thus, P P b Ab td Single shear case where Ab = projected area where bearing pressure is applied P = bearing force Read section 1.8 in text for a detailed stress analysis of a structure. Shearing and Bearing Stress 23 Would like to determine the stresses in the members and connections of the structure shown. From a statics analysis: FAB = 40 kN (compression) FBC = 50 kN (tension) Must consider maximum normal stresses in AB and BC, and the shearing stress and bearing stress at each pinned connection Rod & Boom Normal Stresses The rod is in tension with an axial force of 50 kN. At the rod center, the average normal stress in the circular cross-section (A = 314x10-6m2) is BC = +159 MPa. At the flattened rod ends, the smallest cross-sectional area occurs at the pin centerline, A 20 mm 40 mm 25 mm 300 10 6 m 2 P 50 103 N BC ,end 167 MPa A 300 10 6 m 2 The boom is in compression with an axial force of 40 kN and average normal stress of –26.7 MPa. The minimum area sections at the boom ends are unstressed since the boom is in compression. Pin Shearing Stresses The cross-sectional area for pins at A, B, and C, 2 25 mm 6 2 A r 491 10 m 2 2 The force on the pin at C is equal to the force exerted by the rod BC, P 50 103 N C , ave 102 MPa A 491 10 6 m 2 The pin at A is in double shear with a total force equal to the force exerted by the boom AB, A,ave P 20 kN 40.7 MPa A 491 10 6 m 2 Pin Shearing Stresses Divide the pin at B into sections to determine the section with the largest shear force, PE 15 kN PG 25 kN (largest) Evaluate the corresponding average shearing stress, B,ave PG 25 kN 50.9 MPa A 491 10 6 m2 Pin Bearing Stresses To determine the bearing stress at A in the boom AB, we have t = 30 mm and d = 25 mm, b P 40 kN 53.3 MPa td 30 mm 25 mm To determine the bearing stress at A in the bracket, we have t = 2(25 mm) = 50 mm and d = 25 mm, b P 40 kN 32.0 MPa td 50 mm 25 mm Equilibrium of Shear Stresses Consider an infinitesimal element of material. Apply a single shear stress, τ1. Total shear force on surface is (τ1)bc. 2 1 For equilibrium in the y-direction, apply τ1 on (-) surface. For moment equilibrium about the z-axis, apply τ2 on top and bottom surfaces. Moment equilibrium equation about z-axis: Thus, a shear stress must be balanced by three other stresses for the element to be in equilibrium. 1 2 1 (bc)a 2 (ac)b 1 2 Shearing and Bearing Stress 29 Equilibrium of Shear Stresses What does this tell us? Shear stresses on opposite (parallel) faces of an element are equal in magnitude and opposite in direction. Shear stress on adjacent (perpendicular) faces of an element are equal in magnitude and both point towards or away from each other. Sign convention for shear stresses Positive face – normal is in (+) x, y, or z direction Negative face - normal is in (-) x, y, or z direction Face Direction Shear Stress + + + + - - - - + - + Shearing and Bearing Stress 2 1 - - + 1 + 2 30 Define General State of Stress y z x σx = stress in x-direction applied in the plane normal to x-axis σ y = stress in y-direction applied in the plane normal to y-axis σ z = stress in z-direction applied in the plane normal to z-axis τxy = stress in y-direction applied in the plane normal to x-axis τ xz = stress in z-direction applied in the plane normal to x-axis τ zy = stress in y-direction applied in the plane normal to z-axis And so on… Shearing and Bearing Stress 31 Define General State of Stress y z x There are 9 components of stress: σx, σ y, σ z, τxy, τ xz, τ yx, τ yz, τ zx, τ zy As shown previously, in order to maintain equilibrium: τ xy= τ yx, τ xz = τ zx, τ yz = τ zy There are only 6 independent stress components. Shearing and Bearing Stress 32 Example Problem A load P = 10 kips is applied to a rod supported as shown by a plate with a 0.6 in. diameter hole. Determine the shear stress in the rod and the plate. Shearing and Bearing Stress 33 Example Problem Link AB is used to support the end of a horizontal beam. If link AB is subject to a 10 kips compressive force determine the normal and bearing stress in the link and the shear stress in each of the pins. b 2 in t 1 4 in d 1 in Shearing and Bearing Stress 34 Oblique Planes and Design Considerations (1.11, 1.13) MAE 314 – Solid Mechanics Yun Jing Oblique Planes and Design Considerations 35 Stress on an Oblique Plane What have we learned so far? Axial forces in a two-force member cause normal stresses. Transverse forces exerted on bolts and pins cause shearing stresses. Oblique Planes and Design Considerations 36 Stress on an Oblique Plane Axial forces cause both normal and shearing stresses on planes which are not perpendicular to the axis. Consider an inclined section of a uniaxial bar. The resultant force in the axial direction must equal P to satisfy equilibrium. The force can be resolved into components perpendicular to the section, F, and parallel to the section, V. F P cos V P sin The area of the section is A0 A cos A A0 / cos Oblique Planes and Design Considerations 37 Stress on an Oblique Plane We can formulate the average normal stress on the section as F P cos P cos2 A A0 / cos A0 The average shear stress on the section is V P sin P sin cos A A0 / cos A0 Thus, a normal force applied to a bar on an inclined section produces a combination of shear and normal stresses. Oblique Planes and Design Considerations 38 Stress on an Oblique Plane Since σ and τ are functions of sine and cosine, we know the maximum and minimum values will occur at θ = 00, 450, and 900. P sin cos A0 P cos2 A0 At θ=±900 σ=0 At θ=±900 τ=0 At θ=±450 σ=P/2A0 At θ=±450 τ=P/2A0 (max) At θ=00 σ=P/A0 (max) At θ=00 τ=0 Oblique Planes and Design Considerations 39 Stress on an Oblique Plane What does this mean in reality? Oblique Planes and Design Considerations Design Considerations From a design perspective, it is important to know the largest load which a material can hold before failing. This load is called the ultimate load, Pu. Ultimate normal stress is denoted as σu and ultimate shear stress is denoted as τu. Pu u A Pu u A Oblique Planes and Design Considerations 41 Design Considerations Often the allowable load is considerably smaller than the ultimate load. It is a common design practice to use factor of safety. F.S. ultimateload allowableload F .S. u allowable stress all ultimatestress F .S. Pu Pall ultimate stress allowable stress Oblique Planes and Design Considerations u all 42 Example Problem Two wooden members are spliced as shown. If the maximum allowable tensile stress in the splice is 75 psi, determine the largest load that can be safely supported and the shearing stress in the splice. Oblique Planes and Design Considerations 43 Example Problem A load is supported by a steel pin inserted into a hanging wooden piece. Given the information below, determine the load P if an overall factor of safety of 3.2 is desired. u _ wood 60 MPa (in tension) u _ wood 7.5 MPa u _ steel 145 MPa b 40 m m c 55 m m d 12 m m Oblique Planes and Design Considerations 44