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Hello everyone.
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In the present problem, the diagram of a beam is given as shown.
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It has a roller support at point a and a fixed support at point b.
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The load applied at b, c and a continuous load from d to e is as shown.
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Further, we have a is equal to 0 .4 meter.
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B is 0 .4 meter.
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P is 835 newton.
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W is 2 ,534 newton per meter.
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We have to find bending moment at point b, draw the shear force diagram and also draw the bending moment diagram.
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Coming to the solution, let us first look into the free body diagram of the given problem.
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From the diagram we can see that the force acting at a upwards is ra and the force acting at d upwards is rt, the loads and distances are as shown.
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Coming to the solution further, we have taking moment about a, we get p multiplied by 0 .4 plus p multiplied by 0 .8 plus w multiplied by 0 .4 multiplied by 1 .4 is equal to rt multiplied by 1 .2.
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Therefore, rt is equal to p multiplied by 1 .2 plus 0 .56w divided by 1 .2.
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R .d.
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Is equal to 835 multiplied by 1 .2 plus 0 .56 multiplied by 225 multiplied by 2 .0 .56 multiplied by 2 ,0754 divided by 1 .2.
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Hence rd is equal to 2017 .53 newton, taking moment about d.
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We get ra multiplied by 1 .2 minus p multiplied by 0 .8 minus p multiplied by 0 .8 minus p multiplied by 0 .4 plus w multiplied by 0 .4 multiplied by 0 .0 .0.
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R .a.
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Is equal to p multiplied by 1 .2 minus w multiplied by 0 .08 divided by 1 .2.
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Next, ra is equal to 835 multiplied by 1 .2 minus 2 ,25 multiplied by 2 ,534 multiplied by 0 .08, divided by 1 .2.
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R .a.
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Is equal to 66 .067 newton.
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Next, we have shear force at a is f .a is equal to plus 66 .66 .067 newton.
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At b, fb is equal to 66 .067 minus 835 which is equal to minus 168 .933 newton.
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At c, fc is equal to minus 168 .933 minus 835 which is equal to minus 1 ,033.
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Newton.
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At d, fd is equal to minus 1003 .933 plus 2017 .53, which is equal to 1013 .597.
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At e.
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F .e is equal to 1 ,013 .597 minus 2 ,534 multiplied by 0 .4, which is equal to approximately 0.
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Let us look into the shear force diagram for the given problem.
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Thus, we have the share force diagram as shown.
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So at a, we have 66 .66 .067 newton at b minus 168 .433 newton at c minus 1 ,03 .933 newton at d, 1 ,013 .597 newton and at e 0 newton.
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Further, bending moment at a, m .a...