M (kN.m) e. V (kN) 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 P₁ ولا 25 mm Neutral Axis 200 mm A x D P2 M Support Material 50 mm STRESS FORMULATION 25 mm N 2 Normal stress: σN= Beam's cross-section M₂y L (m) Bending stress: Τρ Torsional shear stress: T= d. Draw the shear force diagram for the beam. Show all your work. V, Q Transverse shear stress: T= Q=YA' or Q= P W 1₂t BEAM SIGN CONVENTION V V ty w L(m) Neutral Axis Pz M (w): Positive internal shear M M Positive internal moment Beam sign convention CROSS-SECTIONAL MOMENTS OF INERTIA ABOUT THE CENTROID y bh³ Quadratic (second) 1, = l₁ = moment of inertia 12 hb3 12 πρ 1₂ = ly=4 Polar moment of inertia 1x = bh3+hb3 12 πρ Draw the bending moment diagram for the beam. Show all your work. P₁ Centroid location: Parallel axis theorem: Σχιλι = Al 1c ==Σ (1c₁ + A₁ d₁²) A L 42 D L (m) Neutral Axis P₂ M x (m)
M (kN.m) e. V (kN) 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 P₁ ولا 25 mm Neutral Axis 200 mm A x D P2 M Support Material 50 mm STRESS FORMULATION 25 mm N 2 Normal stress: σN= Beam's cross-section M₂y L (m) Bending stress: Τρ Torsional shear stress: T= d. Draw the shear force diagram for the beam. Show all your work. V, Q Transverse shear stress: T= Q=YA' or Q= P W 1₂t BEAM SIGN CONVENTION V V ty w L(m) Neutral Axis Pz M (w): Positive internal shear M M Positive internal moment Beam sign convention CROSS-SECTIONAL MOMENTS OF INERTIA ABOUT THE CENTROID y bh³ Quadratic (second) 1, = l₁ = moment of inertia 12 hb3 12 πρ 1₂ = ly=4 Polar moment of inertia 1x = bh3+hb3 12 πρ Draw the bending moment diagram for the beam. Show all your work. P₁ Centroid location: Parallel axis theorem: Σχιλι = Al 1c ==Σ (1c₁ + A₁ d₁²) A L 42 D L (m) Neutral Axis P₂ M x (m)
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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