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M Vable Mechanics of Materials: Chapter 10 Design and Failure Learning objectives Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm • Learn the computation of stresses and strains in structural members subjected to combined axial, torsion, and bending loads • Develop the analysis skills for computation of internal forces and moments on individual members that compromise a structure August 2012 10-1 M Vable Mechanics of Materials: Chapter 10 Combined Loading Axial Non-zero Stresses Non-zero Strains N σ xx = -A σ xx ε xx = -E νσ xx ε yy = – ⎛ -⎞ ⎝ E ⎠ Torsion Tρ τ xθ = J τ xθ γ xθ = G Symmetric Bending about z-axis Mzy σ xx = – ⎛ ⎞ ⎝ I zz ⎠ σ xx ε xx = -E Vy Qz τ xs = – ⎛ -⎞ ⎝ I zz t ⎠ νσ xx ε yy = – ⎛ -⎞ ⎝ E ⎠ My z σ xx = – ⎛ ⎞ ⎝ I yy ⎠ σ xx ε xx = -E Symmetric Bending about y-axis νσ xx ε zz = – ⎛ -⎞ ⎝ E ⎠ τ xs γ xs = G νσ xx ε yy = – ⎛ -⎞ ⎝ E ⎠ τ xs γ xs = G Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Vz Qy τ xs = – ⎛ -⎞ ⎝ I yy t ⎠ August 2012 νσ xx ε zz = – ⎛ -⎞ ⎝ E ⎠ 10-2 νσ xx ε zz = – ⎛ -⎞ ⎝ E ⎠ M Vable Mechanics of Materials: Chapter 10 Combined Axial and Torsional Loading Axial Loading Torsional Loading y D x z D τtor D C A A Free Surface y σaxial C σaxial x z B F S D C A σaxial Free Surface Px A B σaxial τtor Free Surface B B τtor Combined Axial and Torsional Loading y D τtor x z D C A A τtor σaxial σaxial τtor σ C axial T B Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Px B August 2012 10-3 τtor σaxial T τtor C M Vable Mechanics of Materials: Chapter 10 Combined Axial, Torsional and Bending about z-axis Combined Axial and Torsional Loading y D τtor Bending about z-axis Free Surface y A σbend-z x C τtor σaxial C Px B τ C bend-z τtor σaxial B A τbend-z Free Surface Free Surface B σbend-z Py Combined Axial, Torsional and Bending about z-axis y D τtor x z Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm τbend-z-τtor A σaxial−σbend-z D τbend-z+τtor T C A C B σaxial Px τtor B σaxial+σbend-z August 2012 10-4 τtor σ C axial T B Free Surface D A D A D z x z σaxial Py σaxial M Vable Mechanics of Materials: Chapter 10 Combined Axial, Torsional and Bending about y and z-axis Combined Axial, Torsional and Bending about z-axis y D τtor D x z C A Bending about y-axis A y D x z σbend-y τbend-z-τtor σaxial τbend-z+τto C σaxial T B Px τbend-y τtor B P σaxial+σbend-z y D C A A σaxial−σbend-z C σbend-y B Pz B τbend-y Combined Axial, Torsional and Bending about y and z-axis y τtor - τbend-y x z D D C A Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm σaxial−σbend-z τbend-z+τtor C σaxial-σbend-y T τbend-z-τtor A B σaxial+σbend-y Pz Px τtor + τbend-y B P σaxial+σbend-z y August 2012 10-5 M Vable Mechanics of Materials: Chapter 10 Important Points • • • • • • Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm • The complexity of finding the state of stress under combined loading can be simplified by first determining the state of stress due to individual loading Superposition of stresses implies that a stress component at a specific point resulting from one loading can be added or subtracted to a similar stress component from another loading Stress components at different points cannot be added or subtracted nor can stress components which act on different planes or in different directions be added or subtracted The stress formulas give both the correct magnitude correct direction for each stress component if the internal forces and moments are drawn on the free body diagrams as per the sign conventions In a structure, the structural members will have different orientations In order to use subscripts to determine the direction (sign) of stress components, a local x, y, z coordinate system can be established for a structural member such that the x direction is normal to cross-section, i.e., the x-direction is along the axis of the structural member Stresses σyy and σzz are zero for the four load cases Additional stress components can be deduced to be zero by identifying free surfaces The state of stress in combined loading should be shown on a stress cube before processing the stresses for purpose of stress or strain transformation The strains at a point can be obtained from the superposed stress values using the Generalized Hooke’s Law As the normal stresses σyy and σzz are always zero in our structural members, the non-zero strains εyy and εzz are due to the Poisson’s effect, i.e., August 2012 ε yy = ε zz = – νε xx 10-6 M Vable Mechanics of Materials: Chapter 10 Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm General Procedure for Combined Loading • Identify the relevant equations for the problem and use the equations as a check list for the quantities that must be calculated • Calculate the relevant geometric properties (A, Iyy, Izz, J) of the crosssection containing the points where stresses have to be found • At points where shear stress due to bending is to be found, draw a line perpendicular to the center-line through the point and calculate the first moments of the area (Qy, Qz) between free surface and the drawn line Record the s-direction from the free surface towards the point where stress is being calculated • Make an imaginary cut through the cross-section and draw the free body diagram On the free body diagram draw the internal forces and moments as per our sign conventions if subscripts are to be used in determining the direction of stress components Using equilibrium equations to calculate the internal forces and moments • Using the equations identified, calculate the individual stress components due to each loading Draw the torsional shear stress τxθ and bending shear stress τxs on a stress cube using subscripts or by inspection By examining the shear stresses in x, y, z coordinate system obtain τxy and τxz with proper sign • Superpose the stress components to obtain the total stress components at a point • Show the calculated stresses on a stress cube • Interpret the stresses shown on the stress cube in the x, y, z coordinate system before processing these stresses for the purpose of stress or strain transformation August 2012 10-7 M Vable Mechanics of Materials: Chapter 10 C10.1 A solid shaft of inch diameter is loaded as shown in Fig C10.1 The shaft material has a Modulus of Elasticity of E = 30,000 ksi and a Poisson’s ratio of ν = 0.3 The strain gages mounted on the surface of the shaft recorded the following strain values: εa = 2078 μ εb = –1410μ Determine the axial force P and the torque T T a 30⬚ P Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Fig C10.1 August 2012 10-8 b 60⬚ M Vable Mechanics of Materials: Chapter 10 C10.2 A rectangular hollow member is constructed from a 1/2 inch thick sheet metal and loaded as shown Fig C10.2 Determine the normal and shear stresses at points A and B and show it on the stress cubes for P1 = 72 kips, P2 = 0, and P3 = kips y in in A P1 B z 60 in Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Fig C10.2 August 2012 10-9 P3 P2 M Vable Mechanics of Materials: Chapter 10 C10.3 A load is applied to bent pipes as shown By inspection determine and show the total stresses at points A and B on stress cubes using the following notation for the magnitude of stress components: σaxial —axial normal stress; τtor—torsional shear stress; σbend-y —normal stress due to bending about y-axis; τbend-y —shear stress due to bending about y-axis; σbend-z —normal stress due to bending about z-axis; τbend-z —shear stress due to bending about z-axis y A x B z c a b Pz Class Problem y x Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm z A B Py a August 2012 10-10 b M Vable Mechanics of Materials: Chapter 10 C10.4 A pipe with an outside diameter of 40 mm and wall thickness of 10 mm is loaded as shown in Fig C10.5 Determine the normal and shear stresses at point A and B in the x, y, and z coordinate system and show it on a stress cube Points A and B are on the surface of the pipe.Use a = 0.25 m, b = 0.4 m, and c = 0.1 m C10.5 Determine the maximum normal stress and maximum shear stress at point B on the pipe shown in Fig C10.5 y A x B z P = 10 kN a b Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm Fig C10.5 August 2012 10-11 15o c M Vable Mechanics of Materials: Chapter 10 C10.6 The bars in the pin connected structure shown are circular bars of diameters that are available in increments of mm The allowable shear stress in the bars is 90 MPa Determine the diameters of the bars for designing the lightest structure to support a force of P = 40 kN C B P 0.5 m mm 1.6 D A Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm 1.2 m August 2012 10-12 M Vable Mechanics of Materials: Chapter 10 C10.7 A hoist is to be designed for lifting a maximum weight of W = 300 lbs The hoist will be installed at a certain height above ground and will be constructed using lumber and assembled using steel bolts The lumber rectangular cross-section dimensions are listed in table below The bolt joints will be modeled as pins in single shear Same size bolts will be used in all joints The allowable normal stress in the wood is 1.2 ksi and the allowable shear stress in bolts is ksi Design the lightest hoist by choosing the lumber from the given table and bolt size to the nearest 1/8 inch diameter Cross-section Dimension in x in in x in ft in x in ft ft in x in ft P in x in W in x in ft in x in in x in Printed from: http://www.me.mtu.edu/~mavable/MoM2nd.htm in x in August 2012 10-13 ... August 2012 10- 10 b M Vable Mechanics of Materials: Chapter 10 C10.4 A pipe with an outside diameter of 40 mm and wall thickness of 10 mm is loaded as shown in Fig C10.5 Determine the normal and shear... Fig C10.2 August 2012 10- 9 P3 P2 M Vable Mechanics of Materials: Chapter 10 C10.3 A load is applied to bent pipes as shown By inspection determine and show the total stresses at points A and B... for the purpose of stress or strain transformation August 2012 10- 7 M Vable Mechanics of Materials: Chapter 10 C10.1 A solid shaft of inch diameter is loaded as shown in Fig C10.1 The shaft material