2. Conclusions on Transverse Shearing Stress Calculations Maximum Value at Neutral Axis – IF CROSS- SECTION IS NOT NARROWER ELSEWHERE –Depends on Shape of Cross Section (Table 4.3) –Equal to Zero at Top and Bottom Boundaries Important for Short Beams –Wood Beams – if span/depth < 24 –Metal Beams with Thin Webs- if span/depth<15 –Metal Beams with Solid Section-if span/depth<8

3. Summary of Stress Analysis Procedure in Symmetrical Beams Find Centroid of Crosss- Section Neutral Axis (Table 4.2) Choose x-y-z Ref. System Centered at Centroid Normal Bending Stress –Zero at neutral axis –Max. at top and/or bottom surface Transverse Shear Stress –Zero at top and bottom edge –Max. at neutral axis unless cross- section is narrower elsewhere –Z(y=y 1 )=width of cross-section where shear stress is calculated yA = Moment of area above y=y 1 with respect to neutral axis

4. Summary of Solutions to Textbook Problems Problem 4-7: Cantilever beam subjected to biaxial bending. Use superposition of maximum stresses at critical point: –Max. Bending Stress =26,122+22,857=48,979psi –Expect failure by yielding since S Y =43,000 psi for AISI 1020 annealed carbon steel (Table 3.3) Problem 4-8: Beam with fixed ends subjected to concentrated force at mid-span, and circular cross-section of diameter equal to 1 in. –Check for possible failure by yielding, using Case 6 in Table 4.1, to obtain:  max =(600)(0.5)/0.0491=6110 psi, which is smaller than S Y =51,000psi for AISI 1020 cold drawn steel  no yielding failure –Check for failure due to force-induced deformation, using Table 4.1 to obtain: y max =0.0098>0.005inches  design is NOT approved

6. Summary of Solutions to Textbook Problems (Continued) Problem 4-9: Thin plate subjected to tension and temperature change –Check for failure by yielding at max. temperature: not predicted since  max =17,500/[(0.0625)(3)]=93,333psi, whereas S Y =101,000 psi for Ti-6Al-4V alloy at 400 0 F (see Table 3.5) –Check for failure by force-induced deformation: NOT predicted since  f =  f L 0 =(93,333/16x10 6 )(20)=0.1167”<  cr =0.1250”, where E=16x10 6 psi is found for Titanium in Table 3.9 –Check for failure by temperature induced elastic deformation: NOT predicted since  t =L 0  T=(20)(5.3x10 -6 )(325)=0.0345”<0.1250”, where the CTE, , is found from Table 3.8 for the Ti-6Al-4V alloy. –Failure OCCURS due to superposition of  f +  t =0.1512>specified limit of 0.125” Problem 4-10: Submit solution for yielding failure only, i.e., ignore for now the presence of the cracks

11. Summary of Solutions to Textbook Problems (Continued) Problem 4-14: Cylindrical boss fixed at one end and loaded by force “F” at free end –Critical points for bending and transverse shear –Infinitesimal volume elements at critical points –Calculate maximum stress components,  x =347 Mpa, and  xz =  zx =(4/3)(F/A)=173 Mpa, according to Table 4.3 for solid circular section –No failure by yielding for uniaxial tensile stress, since S yp =351.7 Mpa for cold-rolled AISI steel in Table 3.3 –Distortion energy theory of failure is required for the multi- axial state of stress associated with transverse shear Problem 4-15: Submit solution ONLY for failure modes associated with uniaxial state of stress