L17_ Eccentric Axial Loading- fall10

L17_ Eccentric Axial Loading- fall10 - EAS 209-Fall 2010...

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EAS 209-Fall 2010 Instructors: Christine Human 9/17/2010 1 Lecture 17- Eccentric Axial Loading Lecture 17 Eccentric Axial Loading Centric Loading – Line of action of load P passes through the centroid of the section. Produces uniform normal stress distribution in the section. Eccentric Loading – Line of action of P does not pass through the centroid of the section. Today’s Objective: Calculate stresses due to eccentric loading Today’s Homework: e e EAS 209-Fall 2010 Instructors: Christine Human 9/17/2010 2 Lecture 17- Eccentric Axial Loading In order for segment AC to satisfy equilibrium ( Σ M A ), the internal forces at C require a centric load F=P and a moment M=Pe . Stress due to eccentric loading can be found by superposition of the uniform stress due to a centric load and the linear stress distribution due a pure bending moment y o ( ) ( ) I My A P x x x = + = bending centric σ
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EAS 209-Fall 2010 Instructors: Christine Human 9/17/2010 3 Lecture 17- Eccentric Axial Loading The neutral axis corresponds to the point where σ x =0. Ae I y MA PI y I My A P o o o = = = 0 For centric loads, e =0 and y 0 = , uniform stress distribution. As e increases, y 0 decreases and moves towards the centroidal axis. For pure bending, y 0 =0 y o is the distance between the neutral axis and the centroidal axis I is the moment of inertia about the centroidal axis Note that M=Pe EAS 209-Fall 2010 Instructors: Christine Human 9/17/2010 4 Lecture 17- Eccentric Axial Loading Example Solution: Find the equivalent centric load and bending moment
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L17_ Eccentric Axial Loading- fall10 - EAS 209-Fall 2010...

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