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MAE 241 - Statics Summer 2009 Dr. Konstantinos A. Sierros Office Hours: M and W 10:30 – 11:30 (263 ESB new add) [email protected] Teaching Blog:

MAE 241-Lec11

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Page 1: MAE 241-Lec11

MAE 241 - Statics

Summer 2009

Dr. Konstantinos A. SierrosOffice Hours: M and W 10:30 – 11:30 (263 ESB new add)

[email protected] Blog: http://wvumechanicsonline.blogspot.com

Page 2: MAE 241-Lec11

Determine the internal normal force, shear force, and moment at point C in the simply supported beam. Point C is located just to the right of the 1500-lb –ft couple moment.

Page 3: MAE 241-Lec11

Determine the internal normal force, shear force, and moment at point C in the simply supported beam.

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Draw the shear and moment diagrams for the simply supported beam.

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7.3 Relations between distributed load, shear and moment

Distributed loading• Distributed loading is positive when the loading acts upward• The internal shear force and bending moment are assumed to act in the positive sense according to the established sign convention• Distributed loading has been replaced by a resultant force ΔF that acts at distance k Δx

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7.3 Relations between distributed load and shear

)(xwdx

dV

Slope of shear diagram Distributed load intensity

Page 7: MAE 241-Lec11

7.3 Relations between distributed load and shear

Change in shear Area under loading curve

dxxwV )(

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7.3 Relations between the shear and moment

Vdx

dM

Slope of momentdiagram

Shear

VdxM

Change in moment Area under shear

diagram

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7.3 Couple moment

Letting Δx→0

Moment equilibrium requires;

ΣM = 0 ; ΔM = Mo+

The ‘jump’ in the moment diagram isupward if Mo is clockwise.

ΔM is downward when Mo is counterclockwise.

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7.4 Cables

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7.4 Cables

• Flexible cables and chains combine strength with lightness• Assume cable is perfectly flexible and inextensible• The cable takes a form of several straight line segments• Use equations of equilibrium and geometry