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Overview of Loads ON and IN Structures / Machines

Overview of Loads ON and IN Structures / Machines. Overview of Various Stress Patterns. Solution of Example Problem 4.3. Step 1 – Construct shear&moment diagrams Step 2 – Find d min =0.90in to resist  allowable =35,000psi at section where M max =2500in-lb

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Overview of Loads ON and IN Structures / Machines

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  1. Overview of Loads ON and IN Structures / Machines

  2. Overview of Various Stress Patterns

  3. Solution of Example Problem 4.3 • Step 1 –Construct shear&moment diagrams • Step 2 –Find dmin=0.90in to resist allowable=35,000psi at section where Mmax=2500in-lb • Step 3 –Find dmin=0.56in to resist allowable=20,200psi for maximum direct average shear stress • Step 4 –Find dmin=0.65in to resist allowable=20,000psi for maximum transverse shear stress

  4. Example 4.2- Calculation of Transverse Shear Stresses • Hollow rectangular cross-section (Channel) • Moment of Inertia about neutral axis, Izz=8.42 in4 • Max. transverse shear stress at neutral axis • Direct application of “area moment” method • xy(max)=0.34V, xy(ave)=0.2V, so that • xy(max) =1.7 xy(ave)=1.7(F/A)….Table 4.3 • Divide irregular section into several regular parts • Transverse shearing stress distribution, xy(y)

  5. Summary of Solutions to Textbook Problems – Problem 4.10 • Module support D6AC steel beam with two small hole and cracks • Ignore for now stress concentrations&fracture mechanics • Beam subjected to four-point bending • Uniform bending moment at mid-span, M=PL/3 =281.25 kN-m • No transverse shear stress in central span, V =0 • Check for possible failure by yielding • Max. bending stress at: • The tip of bottom crack, yct =10.3cm, x(ct)=(M*yct)/Izz=556 MPa • At the outer fibers: x(max)=(281.25x103)(12.5x10-2)/5.21x10-5=675 Mpa • From Table 2.1, Syp=1570MPa, so that the “safety factor” is equal to 1570/675=2.3  NO yielding

  6. Summary of Solutions to Textbook Problems – Problem 4.15 • Short tubular cantilever bracket, AISI 1020 steel (CD) • Critical points at the wall: • Top and bottom fibers for bending • At neutral axis for transverse shearing stress • Calculate maximum stresses • Axial bending stress: x=48,735 psi • Transverse shear: use Table 4.3 to obtain yz(ts)=2(F/A)=48,890psi • Yielding failure mode for uniaxial tensile stress • No failure since Table 3.3 shows Sy=51,000psi>48,735 psi • Multiaxial failure theory is necessary for yz(ts)

  7. Example 4.5- Calculation of Stresses in Channel-Section Cantilever Beam • Determine maximum bending stress • z(max)=48000(2.0)/4.74 =20,250 psi, Ixx=4.74in4 • Max. transverse shear at neutral axis • Area moment methods yields: zy(max)= 7720 psi • Flow of torsional shearing stresses • Locate SHEAR CENTER by using Table 4.5: e=0.56 in • Find torsional moment since the plane of “P” is located at a distance a=1.26 in. from the shear center: T=Pa =10,080 in-lb • Find horizontal shear forces that form resisting couple, T = Rd • Maximum torsional shearing stresses in the flanges: • zx(max)= R/Af= 2739/((1.72)(0.32)) = 4976 psi • Could be eliminated by translating the plane of “P” by 1.26in to the left

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