Chap1 slides (2)

M Vable Mechanics of Materials: Chapter Stress • A variable that can be used as a measure of strength of a structural member Learning objectives • Understanding the concept of stress • Understanding the two step analysis of relating stresses to external forces and moments Static equivalency Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Equilibrium August 2012 1-1 M Vable Mechanics of Materials: Chapter Normal Stress Tensile Normal Stress Tensile Normal Force N σavg Imaginary Cut Chandelier Weight Chandelier Weight Building Weight Building Weight Imaginary Cut N N σavg N Compressive Normal Force • • • • Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm • σavg σavg Compressive Normal Stress σ av = N ⁄ A All internal forces (and moments) in the book are in bold italics Normal stress that pulls the surface away from the body is called a tensile stress Normal stress that pushes the surface into the body is called a compressive stress The normal stress acting in the direction of the axis of a slender member (rods, cables, bars, columns, etc.) is called the axial stress The compressive normal stress that is produced when one surface presses against other is called the bearing stress Abbreviation August 2012 Units Basic Units psi Pounds per square inch lb/in.2 ksi Kilopounds (kips) per square inch 103 lb/in.2 Pa Pascal N/m2 kPa Kilopascal 103 N/m2 MPa Megapascal 106 N/m2 GPa Gigapascal 109 N/m2 1-2 M Vable Mechanics of Materials: Chapter C1.1 A kg light shown in Fig C1.1 is hanging from the ceiling by wires of diameter of 0.75 mm Determine the tensile stress in the wires AB and BC 2m 2.5 m B 2.5 m A A C Light Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Fig C1.1 August 2012 1-3 M Vable Mechanics of Materials: Chapter Shear Stress Weight of the Clothes Imaginary cut between the wall and the tape Imaginary cut along the possible path of the edge of the ring Pull of the hand Mwall V V Weight of the Clothes V V Mwall τ τ τ Weight of the Clothes τ τ (a) Pull of the hand (b) Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm τ av = V ⁄ A August 2012 Pull of the hand 1-4 M Vable Mechanics of Materials: Chapter Shear stress in pins • Visualizing the surface on which stress acts is very important Single Shear Double Shear F F F F V V V Multiple forces on a pin ND NC NB D C B VB NB Cut B VD Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Cut August 2012 VD VB D C ND NC 1-5 M Vable Mechanics of Materials: Chapter C1.2 The device shown in Fig C1.2 is used for determining the shear strength of the wood The dimensions of the wood block are in x in x 1.5 in If the force required to break the wood block is 12 kips, determine the average shear strength of the wood P in Wood in in Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Fig C1.2 August 2012 1-6 M Vable Mechanics of Materials: Chapter C1.3 Two cast iron pipes are held together by a bolt as shown The outer diameters of the two pipes are 50 mm and 70 mm and wall thickness of each pipe is 10 mm.The diameter of the bolt is 15 mm The bolt broke while transmitting a torque of kN-m On what surface(s) did the bolt break? What was the average shear stress in the bolt on the surface where it broke? T T Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Fig C1.3 August 2012 1-7 M Vable Mechanics of Materials: Chapter C1.4 Fig C1.4 shows a truss and the sequence of assembly of members at pin H All members of the truss have a cross-sectional area of 250 mm2 and all pins have a diameter of 15 mm (a) Determine the axial stresses in members HA, HB, HG and HC of the truss shown in Fig C1.4 (b)Determine the maximum shear stress in pin H G F H 300 B A 3m 3m 3m kN August 2012 E 3m kN Fig C1.4 Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm kN D C 300 1-8 HA HB HG Pin H HC M Vable Mechanics of Materials: Chapter Internally Distributed Force System (a) Normal to plane ␶A ␴A FB A B C FA FD FC A D B E D C E (a) FE Tangent in plane (b) • The intensity of internal distributed forces on an imaginary cut surface of a body is called the stress on a surface • The intensity of internal distributed force that is normal to the surface of an imaginary cut is called the normal stress on a surface • The intensity of internal distributed force that is parallel to the surface of an imaginary cut surface is called the shear stress on the surface • Relating stresses to external forces and moments is a two step process Static equivalency Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Uniform Normal Stress σavg Uniform Shear Stress τavg Equilibrium Normal stress x linear in y y y z z N = σ avg A V = τ avg Uniform shear stre in tangen direction Normal stress linear in z x x A y y z T x My z Mz (a) August 2012 (b) (c) 1-9 (d) (e) M Vable Mechanics of Materials: Chapter C1.5 An adhesively bonded joint in wood is fabricated as shown in Fig C1.5 The joint is to support a force P = 25 kips, what should be the length L of the bonded region if the adhesive strength in shear is 300 psi P in in in L Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Fig C1.5 August 2012 P in 1-10 M Vable Mechanics of Materials: Chapter Class Problem In problems below, draw the free body diagram (FBD) that can be used for calculation of shear stress Identify the surface and the direction of shear stress 1a A nail is being pulled out using a claw hammer Assuming the nail does not bend or break and the hammer does not slip Show the shear stress on the nail in the FBD P 12 in in Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm 1b Two pipes that were adhesively bonded Show the shear stress in the adhesive in the FBD 1c Bolts are used to hold the coupling togather Show the shear stress in the bolts in the FBD August 2012 1-11 P P T T M Vable Mechanics of Materials: Chapter Stress at a Point Outward normal i Internal Force ⎛ ΔF j⎞ σ ij = lim ⎜ -⎟ ΔA i → 0⎝ ΔA i⎠ ΔAi ΔFj direction of outward normal to the imaginary cut surface direction of the internal force component • ΔAi will be considered positive if the outward normal to the surface is in the positive i direction • A stress component is positive if numerator and denominator have the same sign Thus σij is positive if: (1) ΔFj and ΔAi are both positive (2) ΔFj and ΔAi are both negative σ xx τ xy τ xz • Stress Matrix in 3-D: τ yx τ zx σ yy τ yz τ zy σ zz Table 1.1 Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Quantity August 2012 Comparison of number of components One Dimension Two Dimensions Three Dimensions Scalar = 10 = 20 = 30 Vector = 11 = 21 = 31 Stress = 12 = 22 = 32 1-12 M Vable Mechanics of Materials: Chapter Stress Element • Stress element is an imaginary object that helps us visualize stress at a point by constructing surfaces that have outward normal in the coordinate directions Construction of a Stress Element for Axial Stress y (a) (b) y σxx P P x σxx B A z z Stress components are distributed forces on a surface y σxx σxx x Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm z August 2012 1-13 x M Vable Mechanics of Materials: Chapter Construction of a Stress Element for Plane Stress: All stress components on a plane are zero 3-dimensional element y σ xx τ xy τ yx σ yy 0 0 ␶xy B dy A dx ␴yy σyy C ␶yx D z y ␴yy C ␴xx 2-dimensional element σxx ␴xx x τyx τxy B dy A τxy σxx dx dz τyx D σyy Symmetric Shear Stresses: τ xy = τ yx τ yz = τ zy τ zx = τ xz • A pair of symmetric shear stress points towards the corner or away from the corner Stress cube showing all positive stress components τ xy τ xz τ yx σ yy τ yz τ zx τ zy σ zz Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm σ xx August 2012 1-14 x M Vable Mechanics of Materials: Chapter C1.6 Show the non-zero stress components on the A,B, and C faces of the cube σ xx = 90MPa ( C ) τ xy = – 200 MPa τ xz = τ yx = – 200 MPa σ yy = 175MPa ( C ) τ yz = 225MPa τ zx = τ zy = 225MPa σ zz = 150MPa ( T ) x C B z A y Class Problem Show the non-zero stress components of problem C1.6 on the A,B, and C faces of the cube below y z Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm C B x August 2012 A 1-15 M Vable Mechanics of Materials: Chapter C1.7 Show the non-zero stress components in the r, θ, and x cylindrical coordinate system on the A,B, and C faces of the stress element shown σ rr = 145MPa ( C ) τ rθ = 100MPa τ rx = – 125 MPa τ θr = 100MPa σ θθ = 160MPa ( T ) τ θx = 165MPa τ xr = – 125 MPa τ xθ = 165MPa σ xx = 150MPa ( T ) A r B x Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm C August 2012 θ 1-16