Given Position of center of gravity of Rectanglar portion = 17in Position of center of gravity

Feb 3rd, 2012
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Given Position of center of gravity of Rectanglar portion = 17in Position of center of gravity of Rectanglar portion = 18in Position of center of gravity of Rectanglar portion = 18.66in Find Stress, Moments and Moment of Inertia to be calculated in supporting Beams Assumptions 1. All the Beams are Rectangular in shape 2. Weight is uniformly Distributed across all the beams 3. Cantilevered structure is fixed at a point on the ground.

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Cantilevered Roof Structure AnalysisStrength of Materials Problem Written and Solved for MET406ByYour NameOn1 December 2014Problem Statement Analyze Cantilevered Roof Structure problems and predict its Strength and capacityGivenPosition of center of gravity of Rectanglar portion = 17inPosition of center of gravity of Rectanglar portion = 18inPosition of center of gravity of Rectanglar portion = 18.66inFindStress, Moments and Moment of Inertia to be calculated in supporting Beams Assumptions1. All the Beams are Rectangular in shape2. Weight is uniformly Distributed across all the beams 3. Cantilevered structure is fixed at a point on the ground. Method1. Moment of inertia calculated for the rectangular beams considering the position of center of gravity2. Bending moment diagrams and Shear Stress Diagrams are drawn after acute analysis of the structure 3. Stress Distribution calculated for determining the stability of the structureMET406 Topic(s)This Project problem is based on Calculation of Moment of Inertia, Centroid & Stress Distribution in the Mechanical Structure SolutionMoment of Inertia about Y- Axis = M.O.I of square about Y axis + M.O.I of Triangle about Y-Axis - M.O.I of semicircular portion about Y-AxisNow moment of inertia of each geometrical section can be calculated using Parallel axis theorem Iy(Rectangular) = + bh x hcg2 = + 10 x 10 x 172Irect = 29733.33in4Iy (triangular loop) = + bh x hcg2Itria = + 7 x 10 x 18.662Itriangle

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