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# Bending problems part 1

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• 1. Bending Problems Part1 − − − = → = , = → = Prob. 4.1: Knowing that the couple shown acts in a vertical plane,determine the stress at (a) point A. (b) point B.Solution: ℎ3 80 120 3 40 80 3 = = − × 10−12 = 9.8133 × 10−6 4 12 12 12 − = At A: − 0.04 (15000) = = 61.14 ( ) 9.8133 × 10−6
• 2. At B: − −0.06 (15000) = = 91.7 ( ) 9.8133 × 10−6Prob. 4.2: Knowing that the couple shown acts in a vertical plane,determine the stress at (a) point A. (b) point B.Solution:From the symmetry of the cross section, we can see obviouslythat the centroidal axis is z-axis. ℎ3 1 = − 2 4 12 4 (120)(60)3 1 = − 2 (19)4 = 1.12922519 ∗ 10−8 4 12 4 − = At point A: − 0.03 (2000) = = −5.31 1.12922519 ∗ 10−8
• 3. At point B: − −0.019 (2000) = = 3.365 1.12922519 ∗ 10−8Prob. 4.3: A beam of the cross section shown is extruded from analuminum alloy for which = = .Using a factorof safety of 3.00, determine the largest couple that can be appliedto thebeam when it is bent about the z-axis.Solution:
• 4. 450 = = = 150 . 3Since the beam remains in the elastic range and wecan apply our relations mentioned above. ℎ3 16 80 3 16 32 3 = = − × 10−12 = 1.409 × 10−6 4 12 12 12 − ∗ = = − 150 ∗ 106 ∗ 1.409 ∗ 10−6∴ =− = 5.283 . −0.04Prob. 4.5: The steel beam shown is made of a grade of steel forwhich = = . Using a factor of safety of2.50, determine the largest couple that can be appliedto the beamwhen it is bent about the x-axis.Solution: 400 = = = 160 . 2.5
• 5. Since the beam remains in the elastic range and wecan apply our relations mentioned above. 3 3 10 228 200 16 2 = +2∗ + 200 16 122 × 10−12 12 12= 1.0527109 × 10−4 4 ∗ 160 ∗ 106 ∗ 1.0527109 × 10−4 =− =− −0.13= 129.564 . Prob. 4.10: Two equal and opposite couples of magnitude = . are applied to the channel-shaped beam AB.Observing that the couples cause the beam to bend in a horizontalplane, determine the stress at (a) point C, (b) point D. (c) point E.
• 6. Solution:Part # Area Z`i Z`i A 1 4320 18 77760 0.12(0.036)3 = 3.443 ∗ 10−6 m4 + 0.00432(0.02625)2 12 0.03(0.12)3 + 0.0036(15.75 ∗ 10−3 )2 = 5.213 ∗ 10−6 m4 2 3600 60 216000 12 0.03(0.12)3 = 5.213 ∗ 10−6 m4 3 3600 60 216000 + 0.0036(15.75 ∗ 10−3 )2 12 ` 509760 = = = 44.25 11520 = 1 + 2 + 3 = 1.386936 ∗ 10−5 4
• 7. At point (C): − −0.04425 ∗ 25000 = =− = 79.76 () 1.386936 ∗ 10−5At point (D): − 0.07575 ∗ 25000 = =− = −136.54 () 1.386936 ∗ 10−5At point (E): − −0.00825 ∗ 25000 = =− = 14.87 () 1.386936 ∗ 10−5