University:
Kentucky State University
Course:
PHY 211 | General Physics I
Academic year:

2022
Views:

243

Pages:

19
Author:

pravitate8aj15

Lecture 9 Chapter (4) Compression members: * Structural elements that are subjected to only axial compressive forces along longitudinal axis [Image of a column under axial compression] f = A / P Considered uniform stress, not happen in reality Because: eccentricity of the load Type of compression members: 1. Columns: * Vertical member, supported by vertical load. The case of beam-column: chapter 6 2. Trusses 3. Bracing system 4.2 Column theory: [Image of a column buckling under axial load] Pcr: critical buckling load * large load to make it unstable * small load to make it stable P applied slowly → stocky member * Stocky member → may fail by yielding, not buckling Pcr = (π²EI) / L² Euler buckling load E: modulus of elasticity I: moment of inertia L: length of member This equation must valid under 2 conditions: 1. The member must be elastic 2. Its end must be free to rotate but not translate laterally "hinges & pins" [Image of a beam under axial load] M = Pcry y: deflection of the member d²y/dx² = -M/EI y'' + (Pcr/EI)y = 0 2nd order linear equation Example 4.8 W18, DL = 100 kips, LL = 300 kips KL = 26 ft, Fy = 50 ksi 1. Pu = 1.2D + 1.6L = 600 kips 2. Assume Fcr = 0.877Fy = 33 ksi (2/3 Fy) 3. Ag = Pu / (Фc Fcr) = 600 / (0.9)(33) = 20.2 in² 4. Try W18 x 71: Ag = 20.9 in² > 20.2 in² → OK 5. KL/r = (26)(12) / 183.5 = 17 < 200 → OK 6. Fe = π²E / (KL/r)² = π²(29000) / (183.5)² = 8.5 ksi 7. Check: 4.71 √(Fy/E) = 113 < KL = 183.5 → elastic 8. Fcr = 0.877Fe = 7.455 ksi 9. ФcPu = (0.9)(7.455)(20.9) = 140 kips < 600 kips → OK, safe work. See the Example Keep trying until you get a value estimate that Pa < design strength. Lecture 13 4.7 More on effective length * If Compression member is supported differently with respect to each of its principle axes, the effective length will be different for the two directions. Example: Main or minor buckling 15 | | | | 13' | | | 13 beam web column Front view KL = 15 Major or minor buckling 13 | | | | 13 | | | 13 I I I I I I I I I Side view KL = 8 (KL/ry) & (KL/rx) (calculate & take the larger one)