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Column Design ( ) MAE 316 – Strength of Mechanical Components

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Presentation on theme: "Column Design ( ) MAE 316 – Strength of Mechanical Components"— Presentation transcript:

1 Column Design (4.11-4.13) MAE 316 – Strength of Mechanical Components
NC State University Department of Mechanical and Aerospace Engineering Column Design

2 Real world examples Column Design

3 Continue… Jack Column Design

4 Column failure Quebec Bridge designed by Theodore Cooper. It collapsed in 1907. actual weight of the bridge was far in excess of its carrying capacity. Column Design

5 Introduction “Long” columns fail structurally (buckling failure)
“Short” columns fail materially (yielding failure) We will examine buckling failure first, and then transition to yielding failure. P P “Long” “Short” Column Design

6 Buckling of Columns (4.12) For a simply supported beam (pin-pin)
y, v x F.B.D  v(x) The solution to the above O.D.E is: Column Design

7 Buckling of Columns (4.12) Apply boundary conditions: v(0)=0 & v(l)=0
Either A = 0 (trivial solution, no displacement) or The critical load is then The lowest critical load (called the Euler buckling load) occurs when n=1. where n = 1, 2, … Column Design

8 Buckling of Columns (4.12) The corresponding deflection at the critical load is The beam deflects in a half sine wave. The amplitude A of the buckled beam cannot be determined by this analysis. It is finite, but non-linear analysis is required to proceed further Pcr A Column Design

9 Buckling of Columns (4.12) Define the radius of gyration as
Pcr can be written as If l/k is large, the beam is long and slender. If l/k is small, the beam is short an stumpy. To account for different boundary conditions (other than pin- pin), use the end-condition constant, C. Column Design

10 Buckling of Columns (4.12) Theoretical end condition constants for various boundary conditions (Figure 4-18 in textbook) (a) pin-pin (b) fixed-fixed (c) fixed-free (d) fixed-pin Values of C for design are given in Table 4-2 in the textbook Column Design

11 (1) Unstable equilibrium (3) Neutral equilibrium
Buckling of Columns (4.12) 1.Stability (1) Unstable equilibrium (2) Stable equilibrium (3) Neutral equilibrium Column Design

12 Buckling of Columns (4.12) F F Column Design

13 Example 4-16 Column Design

14 Column Design

15 Example Column AB carries a centric load P of magnitude 15 kips. Cables BC and BD are taut and prevent motion of point B in the xz-plane. Using Euler’s formula and a factor of safety of 2.2, and neglecting the tension in the cables, determine the maximum allowable length L. Use E = 29 x 106 psi. Column Design

16 How to prevent buckling?
1) Choose proper cross-sections 2) End conditions 3) Change compression to tension (cable stayed bridges) Column Design

17 Critical Stress (4.12) Plot σcr=Pcr/A vs. l/k
Note: Sy = yield strength = σy Plot σcr=Pcr/A vs. l/k Long column: buckling occurs elastically before the yield stress is reached. Short column: material failure occurs inelastically beyond the yield stress. A “Johnson Curve” can be used to determine the stress for an intermediate column ? No failure Failure Sy Johnson Curve l/k l/k Column Design

18 Critical Stress (4.13) Johnson equation Notice as l/k  0, Pcr/A  Sy
Where Notice as l/k  0, Pcr/A  Sy Column Design

19 Critical Stress (4.13) To find the transition point (l/k)1 between the Johnson and Euler regions Column Design

20 Design of Columns (4.13) Procedure
Calculate (l/k)1 and l/k If l/k ≥ (l/k)1 use Euler’s equation If l/k ≤ (l/k)1 use Johnson’s equation Notice we haven’t included factor of safety yet… Column Design

21 Example 4-19 Column Design

22 Column Design

23 Example Find the required outer diameter, with F.S.=3, for the column shown below. Assume P = 15 kN, L = 50 mm, t = 5 mm, Sy = 372 MPa, E = 207,000 MPa, and the material is 1018 cold-drawn steel. t do Column Design

24 Example Find the maximum allowable force, Fmax, that can be applied without causing pipe to buckle. Assume L = 12 ft, b = 5 ft, do = 2 in, t = 0.5 in, and the material is low carbon steel. F L b t do Column Design

25 Struts or short compression members
Strut - short member loaded in compression If eccentricity exists, maximum stress is at B with axial compression and bending. Note that it is not a function of length If bending deflection is limited to 1 percent of e, then from Eq. (4-44), the limiting slenderness ratio for strut is Column Design

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