Strength of Material Shear Force and Bending Moment Diagram with a simply supported beam with a point load at Mid Point By : Saurabh Kumar Akhouri

Diagram to explain L/2 Ra-W/2 Rb=w/2 Shear force diagram Base Line /2 w/2 W/4.L Rase ine

Analysis : Load Diagram >Figure above shows a beam AB of length L > Simply Supported at the ends A and B > Carrying a point load W at its middle point C >Reaction at the support will be W/2 as the load is acting at the middle point of the beam > Hence Ra-Rb=w/2

Analysis : Shear Force Diagram Take a section X at a distance x from the end A between A and C :Fx= Shear Force at X Mx-Bending Moment at X > Consider left portion of the beam Shear Force will be the resultant force of the reaction at A which is w/2 and +ve >Shear force will be the resultant force of the reaction at which is w/2 and -ve

Analysis : Bending Moment Diagram > Bending Moment at any section between A and C shall be Or, Mx=+w/2x (i) > Bending Moment shall be +ve as per the left section of the beam The moment for all forces are clockwise > Moreover bending moment takes place in such a way that concavity is at the top of the beam > At A, x=0, hence Ma=W/2 x 0-0 > At C, x=L/2, hence Mc=w/2xl/2-WsL/4 > From equation () it is clear B.M. varies according to the straight line law between A and C > Bending Moment at A-0 > Bending Moment at any section between C and B at a distance x from the end A is

Bending Moment Diagram(Contd Analysis Bending Moment Diagram(Contd..) X- W.x .X > At C, X=L/2, Hence Mc=WL/2-WL/21/2-WL/4 > At B, X=L, Hence Mb= WL/2-WL/2-0 Hence Bending Moment at C is WL/4 and it decreases to 0 at B

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Saurabh Kumar Akhouri

B Tech Petroleum Engineering IIT ISM Dhanbad , Post Graduate Diploma in Development Policy IGIDR Mumbai, Post Graduate Diploma in Management

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Chandan Singh Kunwar

2 years ago

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