10 combined loadings 2015

CHAPTER 10: COMBINED LOADINGS 10.1 Unsymmetric Bending 10.2 Eccentric Axial Loadings 10.3 Torsion and Bending 10.4 General Combined Loadings 10.1 UNSYMMETRIC BENDING UNSYMMETRIC BENDING • Analysis of pure bending has been limited to members subjected to bending couples acting in a plane of symmetry • Members remain symmetric and bend in the plane of symmetry • The neutral axis of the cross section coincides with the axis of the couple • Will now consider situations in which the bending couples not act in a plane of symmetry • Cannot assume that the member will bend in the plane of the couples • In general, the neutral axis of the section will not coincide with the axis of the couple 10.1 UNSYMMETRIC BENDING UNSYMMETRIC BENDING •  Fx    x dA      m dA y  c  or   y dA neutral axis passes through centroid Wish to determine the conditions under which the neutral axis of a cross section of arbitrary shape coincides with the axis of the couple as shown • The resultant force and moment from the distribution of elementary forces in the section must satisfy Fx   M y M z  M  applied couple   • M  M z    y   m dA y  c σ I or M  m c  I  I z  moment of inertia defines stress distribution •  M y   z x dA   z   m dA y  c  or   yz dA  I yz  product of inertia couple vector must be directed along a principal centroidal axis 10.1 UNSYMMETRIC BENDING UNSYMMETRIC BENDING Superposition is applied to determine stresses in the most general case of unsymmetric bending • Resolve the couple vector into components along the principle centroidal axes M z  M cos M y  M sin  • Superpose the component stress distributions Mz y M yz x    Iz Iy • Along the neutral axis, M cos y  M sin z Mz y M yz x      Iz Iy Iz Iy tan   y Iz  tan  z Iy 10.1 UNSYMMETRIC BENDING UNSYMMETRIC BENDING x   Mz y M yz  Iz Iy • Along the neutral axis, x    tan   M cos y  M sin z Mz y M yz   Iz Iy Iz Iy y Iz  tan  z Iy • Equation of Neutral Axis M y Iz y z Mz Iy M y Iz y z Mz Iy 10.1 UNSYMMETRIC BENDING UNSYMMETRIC BENDING – SUMMARY Mx z + y Mu  M x  M y 2 x b h z - - + + Mu y x y x Ix + h x My z   - y x Mx - y y My Iy x b • Equation of Neutral Axis (N.A.) M y Ix y x Mx Iy 10.1 UNSYMMETRIC BENDING UNSYMMETRIC BENDING – SUMMARY -  - h x Compression stress + +  max y b Mx x h x - Mu + b z   Ix y M y y Tension stress • Equation of Neutral Axis (N.A.) y Mx z + My Iy x M y Ix y x Mx Iy (Diagram of Equation : y = ax) 10.1 UNSYMMETRIC BENDING EXAMPLE 10.01 SOLUTION: • Resolve the couple vector into components along the principle centroidal axes and calculate the corresponding maximum stresses M z  M cos M y  M sin  • Combine the stresses from the component stress distributions Mzy Myy x    Iz Iy A 1600 lb-in couple is applied to a rectangular wooden beam in a plane • Determine the angle of the neutral forming an angle of 30 deg with the axis vertical Determine (a) the maximum y Iz tan    tan stress in the beam, (b) the angle that the z Iy neutral axis forms with the horizontal plane 10.1 UNSYMMETRIC BENDING EXAMPLE 10.01 • Resolve the couple vector into components and calculate the corresponding maximum stresses M z  1600 lb  in  cos 30  1386 lb  in M y  1600 lb  in sin 30  800 lb  in 1.5 in 3.5 in 3  5.359 in I z  12 3.5 in 1.5 in 3  0.9844 in I y  12 The largest tensile stress due to M z occurs along AB 1  M z y 1386 lb  in 1.75 in    452.6 psi Iz 5.359 in The largest tensile stress due to M z occurs along AD 2  M yz Iy  800 lb  in 0.75 in   609.5 psi 0.9844 in • The largest tensile stress due to the combined loading occurs at A  max  1    452.6  609.5  max  1062 psi 10.1 UNSYMMETRIC BENDING EXAMPLE 10.01 • Determine the angle of the neutral axis Iz 5.359 in tan   tan  tan 30 Iy 0.9844 in  3.143   72.4o 10.2 ECCENTRIC AXIAL LOADINGS ECCENTRIC AXIAL LOADING IN A PLANE OF SYMMETRY • Stress due to eccentric loading found by superposing the uniform stress due to a centric load and linear stress distribution due a pure bending moment  x   x centric   x bending  • Eccentric loading FP M  Pd P M  y A I • Validity requires stresses below proportional limit, deformations have negligible effect on geometry, and stresses not evaluated near points of load application 10.2 ECCENTRIC AXIAL LOADINGS EXAMPLE 10.02 SOLUTION: • Find the equivalent centric load and bending moment • Superpose the uniform stress due to the centric load and the linear stress due to the bending moment • Evaluate the maximum tensile and compressive stresses at the inner and outer edges, respectively, of the superposed stress distribution An open-link chain is obtained by bending low-carbon steel rods into the shape shown For 160 lb load, determine • Find the neutral axis by determining the location where the normal stress (a) maximum tensile and compressive is zero stresses, (b) distance between section centroid and neutral axis 10.2 ECCENTRIC AXIAL LOADINGS EXAMPLE 10.02 • Normal stress due to a centric load A  c   0.25 in 2  0.1963 in P 160 lb 0   A 0.1963 in  815 psi • Equivalent centric load and bending moment P  160 lb M  Pd  160 lb 0.6 in   104 lb  in • Normal stress due to bending moment I  14 c  14  0.254  3.068  103 in Mc 104 lb  in 0.25 in  m   I 068  103 in  8475 psi 10.2 ECCENTRIC AXIAL LOADINGS EXAMPLE 10.02 • Maximum tensile and compressive stresses t  0 m  815  8475 c  0  m  815  8475  t  9260 psi  c  7660 psi • Neutral axis location 0 P My0  A I P I 3.068 103 in y0   815 psi  AM 105 lb  in y0  0.0240 in 10.2 ECCENTRIC AXIAL LOADINGS EXAMPLE 10.03 The largest allowable stresses for the cast iron link are 30 MPa in tension and 120 MPa in compression Determine the largest force P which can be applied to the link SOLUTION: • Determine an equivalent centric load and bending moment • Superpose the stress due to a centric load and the stress due to bending Given: A   103 m Y  0.038 m I  868 109 m • Evaluate the critical loads for the allowable tensile and compressive stresses • The largest allowable load is the smallest of the two critical loads 10.2 ECCENTRIC AXIAL LOADINGS EXAMPLE 10.03 • Determine an equivalent centric and bending loads d  0.038  0.010  0.028 m P  centric load M  Pd  0.028 P  bending moment • Superpose stresses due to centric and bending loads 0.028 P 0.022  377 P P Mc A P    A I 103 868 109 0.028 P 0.022  1559 P P Mc P B    A    A I 103 868 109 A   • Evaluate critical loads for allowable stresses  A  377 P  30 MPa P  79.6 kN  B  1559 P  120 MPa P  79.6 kN • The largest allowable load P  77.0 kN 10.2 ECCENTRIC AXIAL LOADINGS GENERAL CASE OF ECCENTRIC AXIAL LOADING • Consider a straight member subject to equal and opposite eccentric forces • The eccentric force is equivalent to the system of a centric force and two couples P  centric force M y  Pa M z  Pb • By the principle of superposition, the combined stress distribution is P Mz y M yz x    A Iz Iy • If the neutral axis lies on the section, it may be found from My Mz P y z Iz Iy A 10.2 ECCENTRIC AXIAL LOADINGS SUMMARY Mx + + Mu  M x  M y x y y b My h z x - h x z x - - + - + Mu y x y b y Nz z x y h + + + + x x y Mu y 10.2 ECCENTRIC AXIAL LOADINGS Mu  M x  M y SUMMARY h x Mx - + - h My Iy x Mx Ix y My Iy x Nz A + -  Compression stress Mu y  max + x My y b Nz Ix y x b x Mx z   + y h z   + + x + + y z   Nz Mu z y Tension stress A Equation of Neutral Axis (N.A.): (Diagram of Equation : y = ax +b) M y Ix Nz Ix y x Mx Iy A Mx 10.3 TORSION AND BENDING CIRCULAR SHAFTS Mz = M0 P M0 x Mz   J z  max  J  12  c  J  12  c24  c14  max  Mz c J exists on the perimeter of the circular section 10.3 TORSION AND BENDING CIRCULAR SHAFTS Mz = M0  max  max  max  max  max  4 max    Mu z or  max u  max  3 max      v  max  min2  4 max    or  min2  3 max    10.3 TORSION AND BENDING RECTANGULAR SECTIONS  A H G  ( M x M y )  min( M )   max ( M (M x ) x 1 Mx B A H  max (M y )  max G B OR Mz C F  max My (M x M y ) C D E F D E   max (M y )  max  max ( M )   min( M x y) 1 (M x ) y) 10.3 TORSION AND BENDING RECTANGULAR SECTIONS  TB3   12  4 12     S y 2n OR  TB4    3     Sy 3n 10.4 GENERAL COMBINED LOADINGS RECTANGULAR SECTIONS  ( M x M y )  z A H G  ( Nz ) (M x ) 1  z  min( M )   max ( M )   z ( N ( Nz ) x Mx y B A  max H (M y )  z (Nz )  max G B Nz Mz C F My  max (M x M y )  z (Nz ) C D E F  (M y )  max  z D E ( Nz )  max (M x )   (M y )  z (Nz )  max 1 (M x )  z (Nz ) z) 10.4 GENERAL COMBINED LOADINGS RECTANGULAR SECTIONS  TB3   12  4 12     S y 2n OR  TB4    3     Sy 3n