- #1

richard9678

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I don't get this shear force distribution stuff.

In my drawing A is sort of a free body diagram. Load W is dead center of a beam, and of course this force is supported at either end by two reaction forces.with a value of W/2.

B is a shear force diagram. For points < L/2, shear force is positive and > L/2 is negative. I believe this is the right diagram. I have a book that shows the reverse, but I think it is incorrect. I don't get this diagram at all, why the sudden change in shear force direction.

C attempts to show how you would show a shearing force. S is shown as half an arrow and pointing upwards. I'm not certain that the arrow points in the correct direction, but if this arrow shows the situation with the shear force diagram, maybe it does. If B is correct.

As I understand it, the reaction force on the left, where the support is, with a value of W/2 is opposed by a force -W/2. acting opposite, to achieve equilibrium. This gives a sense, near the middle on the left hand side, that S is something of a reaction force to some force of -W/2 above it.

I have the equation, W/2 + S = 0. S comes to W/2, but I'm not sure whether S is a reaction force or an action force. But, in my diagram C, that force looks like a reaction force acting against some force of -W/2 above it.

For some reason I just don't grasp,where at a point > L/2, the shear force diagram flips.

The equation given is W/2 + S - W = 0

S = W -(W/2)

S = -W/2. I don't grasp that. Shear force is -W/2.

I would guess, that S is a reaction force, when it's positive, because it's going upwards as it does for a section. But, S turns negative (for some reason) and then S appears like a action force. Does S turn from one to another? What is changing when shear force S flips?.

If a length to a point was greater than L/2, would the free body diagram look like D, E or G? Or what?

I'm perplexed. Thanks.

In my drawing A is sort of a free body diagram. Load W is dead center of a beam, and of course this force is supported at either end by two reaction forces.with a value of W/2.

B is a shear force diagram. For points < L/2, shear force is positive and > L/2 is negative. I believe this is the right diagram. I have a book that shows the reverse, but I think it is incorrect. I don't get this diagram at all, why the sudden change in shear force direction.

C attempts to show how you would show a shearing force. S is shown as half an arrow and pointing upwards. I'm not certain that the arrow points in the correct direction, but if this arrow shows the situation with the shear force diagram, maybe it does. If B is correct.

As I understand it, the reaction force on the left, where the support is, with a value of W/2 is opposed by a force -W/2. acting opposite, to achieve equilibrium. This gives a sense, near the middle on the left hand side, that S is something of a reaction force to some force of -W/2 above it.

I have the equation, W/2 + S = 0. S comes to W/2, but I'm not sure whether S is a reaction force or an action force. But, in my diagram C, that force looks like a reaction force acting against some force of -W/2 above it.

For some reason I just don't grasp,where at a point > L/2, the shear force diagram flips.

The equation given is W/2 + S - W = 0

S = W -(W/2)

S = -W/2. I don't grasp that. Shear force is -W/2.

I would guess, that S is a reaction force, when it's positive, because it's going upwards as it does for a section. But, S turns negative (for some reason) and then S appears like a action force. Does S turn from one to another? What is changing when shear force S flips?.

If a length to a point was greater than L/2, would the free body diagram look like D, E or G? Or what?

I'm perplexed. Thanks.

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