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Hello everyone.
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In the present problem we have to find a, the support reactions at ping a and roller e.
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B, shear force and bending moment values at the section passing through point b.
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Coming to the solution, we have for the variable load between e and f.
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Equivalent load will be equal to 12 multiplied by 1 kiloton which is equal to 12 kilo which is equal to 12 kilo newton also it will act normally at the center of ef next for the variable load between d and e.
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The equivalent load will be equal to half multiplied by 1, multiplied by 12 kilo newton, which is equal to 6 kilo newton.
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Normally, one third of the length of the load from the larger end, that is 1 divided by 3 meter from the point e.
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Let us look into the free body diagram of the given beam.
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We have the given beam in which the points are a, b, c, d, e and f.
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At a, we have the horizontal reaction as f -a -x and the vertical reaction as f -a -y, and the vertical reaction as f -a -y.
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At b, we have a force acting downwards which is 8 kilo newton.
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And at c, we have a moment in anticlockwise direction whose value is 15 kilo newton meter.
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And between d and e, we have 6 kilo newton acting in downward direction at a distance of 1 divided by 3.
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Meter from the point e and between e and f we have 12 kilouton acting in the downward direction at a distance of 1 by 2 meter from the point e as 2.
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Now the distance between a and b is given as 1 meter.
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The distance between b and c is given as 0.
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0 .5 meter, the distance between c and d is 0 .5 meter.
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The distance between d and e is 1 meter.
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The distance between e and f is 1 meter.
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Next.
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For equilibrium we have sigma fx is equal to 0, which implies that f -a -x is equal to z...