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- 1. Bending Problems Part1<br />ϵ=-yρ -> σ=-yEρ , M=EIρ -> σ=-yMI<br />Prob. 4.1: Knowing that the couple shown acts in a vertical plane, determine the stress at (a) point A. (b) point B.<br />Solution:<br />I=bh312=80120312-4080312×10-12=9.8133×10-6 m4<br />σ=-yMI<br />At A:<br />σ=-0.04(15000)9.8133×10-6=61.14 MPa (Compressive stress)<br />At B:<br />σ=--0.06(15000)9.8133×10-6=91.7 MPa (Tensile stress)<br />Prob. 4.2: Knowing that the couple shown acts in a vertical plane, determine the stress at (a) point A. (b) point B.<br />Solution:<br />From the symmetry of the cross section, we can see obviously that the centroidal axis is z-axis. <br />Iz=bh312-214πr4<br />Iz=(120)(60)312-214π(19)4=1.12922519*10-8 m4<br />σ=-yMI<br />At point A:<br />σ=-0.03(2000)1.12922519*10-8=-5.31 GPa<br />At point B:<br />σ=--0.019(2000)1.12922519*10-8=3.365 GPa<br />Prob. 4.3: A beam of the cross section shown is extruded from an aluminum alloy for which σy=250 MPa and σU=450 MPa.Using a factor of safety of 3.00, determine the largest couple that can be appliedto the beam when it is bent about the z-axis.<br />Solution:<br />σall=σUF.S=4503=150 MPa<br />Since σall<σythe beam remains in the elastic range and we can apply our relations mentioned above.<br />I=bh312=1680312-1632312×10-12=1.409×10-6 m4<br />σall=-yMI Mlargest=-σall*Iymax<br />∴Mlargest=-150*106*1.409*10-6-0.04=5.283 KN.m<br />Prob. 4.5: The steel beam shown is made of a grade of steel for which σy=250 MPa and σU=400 MPa. Using a factor of safety of 2.50, determine the largest couple that can be appliedto the beam when it is bent about the x-axis. <br />Solution:<br /> σall=σUF.S=4002.5=160 MPa<br />Since σall<σythe beam remains in the elastic range and we can apply our relations mentioned above.<br />I=10228312+2*20016312+200161222×10-12<br />=1.0527109×10-4 m4<br />Mlargest=-σall*Iymax=-160*106*1.0527109×10-4-0.13<br />=129.564 KN.m<br />Prob. 4.10: Two equal and opposite couples of magnitude M=25 KN.m are applied to the channel-shaped beam AB. Observing that the couples cause the beam to bend in a horizontal plane, determine the stress at (a) point C, (b) point D. (c) point E. <br />Solution:<br /> <br /> Iy Z`i A Z`i AreaPart #=3.443*10-6m4=5.213*10-6m4=5.213*10-6m40.12(0.036)312+0.00432(0.02625)20.03(0.12)312+0.0036(15.75*10-3)20.03(0.12)312+0.0036(15.75*10-3)2 77760 216000 216000 18 60 60 4320 3600 3600 1 2 3<br />z=Z`iAA=50976011520=44.25 mm<br />Iy=Iy1+Iy2+Iy3=1.386936*10-5 m4<br />At point (C):<br />σ=-zMI=--0.04425*250001.386936*10-5=79.76 MPa (Tensile)<br />At point (D):<br />σ=-zMI=-0.07575*250001.386936*10-5=-136.54 MPa (compressive)<br />At point (E):<br />σ=-zMI=--0.00825*250001.386936*10-5=14.87 MPa (Tensile)<br />

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