EAS209 L18_ Eccentric Axial Loading- spring 2011

EAS209 L18_ Eccentric Axial Loading- spring 2011 - EAS...

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EAS 209-Spring 2011 Instructors: Christine Human 2/7/2011 1 Lecture 18- Eccentric Axial Loading Lecture 18 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 In order for segment AC to satisfy equilibrium ( M A ), the internal forces at C require a centric load F=P and a moment M=Pe . EAS 209-Spring 2011 Instructors: Christine Human 2/7/2011 2 Lecture 18- Eccentric Axial Loading The NA corresponds to the point where σ x =0. Ae I y MA PI y I My A P o o o 0 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 y o is the distance between the neutral axis and the centroidal axis I is the moment of inertia about the centroidal axis because M=Pe
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EAS 209-Spring 2011 Instructors: Christine Human 2/7/2011 3 Lecture 18- Eccentric Axial Loading Example Solution: Find the equivalent centric load and bending moment Superpose the uniform stress due to the centric load and the linear stress due to the bending moment. An open-link chain is obtained
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EAS209 L18_ Eccentric Axial Loading- spring 2011 - EAS...

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