Mae 241-lec11
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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: http://wvumechanicsonline.blogspot.com
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.
Determine the internal normal force, shear force, and moment at point C in the simply supported beam.
Draw the shear and moment diagrams for the simply supported beam.
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moment are assumed to act in the positive sense according to the established sign convention • Distributed loading has been replaced by a resultant force ΔF that acts at distance k Δx
7.3 Relations between distributed load and shear
dV = w(x) dx Slope of shear diagram Distributed load intensity
7.3 Relations between distributed load and shear
∆V = ∫ w( x)dx Change in shear
Area under loading curve
7.3 Relations between the shear and moment
dM =V dx
∆M = ∫ Vdx Change in moment
Slope of moment diagram
Shear
Area under shear diagram
7.3 Couple moment Letting Δx→0 Moment equilibrium requires; +
ΣM = 0 ; ΔM = Mo
The ‘jump’ in the moment diagram is upward if Mo is clockwise. ΔM is downward when Mo is counterclockwise.
7.4 Cables
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
Dr. Konstantinos A. Sierros Office Hours: M and W 10:30 – 11:30 (263 ESB new add) [email protected] Teaching Blog: http://wvumechanicsonline.blogspot.com
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.
Determine the internal normal force, shear force, and moment at point C in the simply supported beam.
Draw the shear and moment diagrams for the simply supported beam.
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
7.3 Relations between distributed load and shear
dV = w(x) dx Slope of shear diagram Distributed load intensity
7.3 Relations between distributed load and shear
∆V = ∫ w( x)dx Change in shear
Area under loading curve
7.3 Relations between the shear and moment
dM =V dx
∆M = ∫ Vdx Change in moment
Slope of moment diagram
Shear
Area under shear diagram
7.3 Couple moment Letting Δx→0 Moment equilibrium requires; +
ΣM = 0 ; ΔM = Mo
The ‘jump’ in the moment diagram is upward if Mo is clockwise. ΔM is downward when Mo is counterclockwise.
7.4 Cables
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
