Machine Design slide 6

# Machine Design slide 6 - With different end conditions Sr...

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Click to edit Master subtitle style 4/17/11 Machine Design ENGR 3220U Week 3, Lecture 2 UOIT, 2011, Winter

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4/17/11 Example 1
4/17/11 Example 2 Figure P4-23 shows a 1-in-dia steel bar supported and subjected to the applied load P=500 lb. Solve for the deflection at the load and the slope at the roller support.

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Long columns with central loading 2 2 l EI P cr π = When P reaches a critical value, the column will be come unstable and bending will develop

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Compression Members Using relationship I = Ak2, where A is cross section area and k is the radius of gyration, equation in last slide becomes: 2 2 2 2 ) / ( r cr S E k l E A P π π = = Where Sr is called slenderness ratio. This ratio, rather than the actual column length, will be used in classifying column according to length categories.
4/17/11 Slenderness Ratio

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Unformatted text preview: With different end conditions, Sr are different. 4/17/11 Slenderness Ratio k l S eff r = l eff, instead of l , is used to calculate Sr. With different end conditions, l eff is given in above table. 4/17/11 Intermediate-length column with E S S S A P r yc yc cr 1 2 2 -= π Columns with Eccentric Loading For columns with eccentric loading, the unit load can be calculated with secant column formula: (3) 4/17/11 Columns with Eccentric Loading Comparison of secant and Euler equations of steel with Sy=40 psi 4/17/11 Struts or Short Columns For short compression members (struts), the critical unit load can be calculated with following equation: And limiting slender ratio is given by: If slenderness ratio is less than limiting Slender ratio, use above equation to calculate critical unit load, otherwise use secant formula (Equation 3) in slide 16. (4)...
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