2. Overview of Loads ON and IN Structures / Machines

3. Overview of Various Stress Patterns

4. 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 Step 3 –Find d min =0.56in to resist  allowable =20,200psi for maximum direct average shear stress Step 4 –Find d min =0.65in to resist  allowable =20,000psi for maximum transverse shear stress

7. Example 4.2- Calculation of Transverse Shear Stresses Hollow rectangular cross-section (Channel) –Moment of Inertia about neutral axis, I zz =8.42 in 4 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)

11. 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, y ct =10.3cm,  x(ct) =(M*y ct )/I zz =556 MPa At the outer fibers:  x(max) =(281.25x10 3 )(12.5x10 -2 )/5.21x10 -5 =675 Mpa –From Table 2.1, S yp =1570MPa, so that the “safety factor” is equal to 1570/675=2.3  NO yielding

15. 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 S y =51,000psi>48,735 psi Multiaxial failure theory is necessary for  yz(ts)

17. 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, I xx =4.74in 4 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/A f = 2739/((1.72)(0.32)) = 4976 psi Could be eliminated by translating the plane of “P” by 1.26in to the left