M (kN.m) e. A cantilevered beam has of length L = 10 m and is subjected to a vertical load P₁ = 50 kN, an axial load P₂ = 100 kN passing through the neutral axis, a moment M = 200 kN.m, and a uniformly distributed load w = 30 kN/m, as shown. The cross-sectional view and dimensions are illustrated below, with point C representing the centroid of the cross-section. 300 mm W A x D W 2 L (m) P₁ ولا 25 mm Neutral Axis 200 mm P2 Support Material 50 mm M STRESS FORMULATION 25 mm N Normal stress: ON= Beam's cross-section M₂y Bending stress: Τρ Torsional shear stress: T= V, Q Transverse shear stress: T = It Q=YA' or Q= Draw the bending moment diagram for the beam. Show all your work. P₁ A L 42 D L (m) Neutral Axis P₂ M x (m) BEAM SIGN CONVENTION V V Positive internal shear M M Positive internal moment Beam sign convention CROSS-SECTIONAL MOMENTS OF INERTIA ABOUT THE CENTROID y bh³ Quadratic (second) 1,= ly = moment of inertia 12 hb³ 12 πρ 1₂ = ly=4 bh3+hb3 Polar moment of inertia 1x = 12 1x = 177 Σχιλι Centroid location: = Al Parallel axis theorem: ==Σ(1c₁ + A₁ d₁²)
M (kN.m) e. A cantilevered beam has of length L = 10 m and is subjected to a vertical load P₁ = 50 kN, an axial load P₂ = 100 kN passing through the neutral axis, a moment M = 200 kN.m, and a uniformly distributed load w = 30 kN/m, as shown. The cross-sectional view and dimensions are illustrated below, with point C representing the centroid of the cross-section. 300 mm W A x D W 2 L (m) P₁ ولا 25 mm Neutral Axis 200 mm P2 Support Material 50 mm M STRESS FORMULATION 25 mm N Normal stress: ON= Beam's cross-section M₂y Bending stress: Τρ Torsional shear stress: T= V, Q Transverse shear stress: T = It Q=YA' or Q= Draw the bending moment diagram for the beam. Show all your work. P₁ A L 42 D L (m) Neutral Axis P₂ M x (m) BEAM SIGN CONVENTION V V Positive internal shear M M Positive internal moment Beam sign convention CROSS-SECTIONAL MOMENTS OF INERTIA ABOUT THE CENTROID y bh³ Quadratic (second) 1,= ly = moment of inertia 12 hb³ 12 πρ 1₂ = ly=4 bh3+hb3 Polar moment of inertia 1x = 12 1x = 177 Σχιλι Centroid location: = Al Parallel axis theorem: ==Σ(1c₁ + A₁ d₁²)
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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