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Pin support, why is there no horizontal reaction force?
Why build a cell lattice tower with square cross-section instead of triangular cross-section?Logic behind location of shear centreBest angle and length for a diagonal wood support for a pull-up barHow to calculate a structure's ultimate loadHow to transfer reaction forces of beam to a weld?Maximum deflection of a beam, fixed in one end and concetrated load at free endCritical Buckling Load for a Spring Supported BarHow to calculate indeterminacy of pin-jointed frame?How to find the reaction forces, moments and the displacement of the fixed beam with a link?Trying to shed some weight on my design, any thoughts?
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I am trying to understand this problem:
If we break down the 10T, they behave like this:
The right end of the beam can move as much as it wants, but the left end is pinned.
Why is there no horizontal reaction at A, if there is a horizontal part of the forces being applied on the beam axis?
I imagine, the force will push one section, and each section will push the next one, until the end, where it will create a reaction, just like the vertical forces are doing?
Thank you a lot.
structures
asked Jun 16 at 12:42
UniversidadDeSevillaEsBasuraUniversidadDeSevillaEsBasura
112 bronze badges
New contributor
UniversidadDeSevillaEsBasura is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
$begingroup$
I am trying to understand this problem:
If we break down the 10T, they behave like this:
The right end of the beam can move as much as it wants, but the left end is pinned.
Why is there no horizontal reaction at A, if there is a horizontal part of the forces being applied on the beam axis?
I imagine, the force will push one section, and each section will push the next one, until the end, where it will create a reaction, just like the vertical forces are doing?
Thank you a lot.
structures
asked Jun 16 at 12:42
UniversidadDeSevillaEsBasuraUniversidadDeSevillaEsBasura
112 bronze badges
New contributor
UniversidadDeSevillaEsBasura is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I am trying to understand this problem:
If we break down the 10T, they behave like this:
The right end of the beam can move as much as it wants, but the left end is pinned.
Why is there no horizontal reaction at A, if there is a horizontal part of the forces being applied on the beam axis?
I imagine, the force will push one section, and each section will push the next one, until the end, where it will create a reaction, just like the vertical forces are doing?
Thank you a lot.
structures
asked Jun 16 at 12:42
UniversidadDeSevillaEsBasuraUniversidadDeSevillaEsBasura
112 bronze badges
New contributor
UniversidadDeSevillaEsBasura is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am trying to understand this problem:
If we break down the 10T, they behave like this:
The right end of the beam can move as much as it wants, but the left end is pinned.
Why is there no horizontal reaction at A, if there is a horizontal part of the forces being applied on the beam axis?
I imagine, the force will push one section, and each section will push the next one, until the end, where it will create a reaction, just like the vertical forces are doing?
Thank you a lot.
structures
structures
asked Jun 16 at 12:42
UniversidadDeSevillaEsBasuraUniversidadDeSevillaEsBasura
112 bronze badges
New contributor
UniversidadDeSevillaEsBasura is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jun 16 at 12:42
UniversidadDeSevillaEsBasuraUniversidadDeSevillaEsBasura
112 bronze badges
New contributor
UniversidadDeSevillaEsBasura is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jun 16 at 12:42
UniversidadDeSevillaEsBasuraUniversidadDeSevillaEsBasura
112 bronze badges
New contributor
UniversidadDeSevillaEsBasura is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jun 16 at 12:42
UniversidadDeSevillaEsBasuraUniversidadDeSevillaEsBasura
112 bronze badges
asked Jun 16 at 12:42
UniversidadDeSevillaEsBasuraUniversidadDeSevillaEsBasura
112 bronze badges
112 bronze badges
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UniversidadDeSevillaEsBasura is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
UniversidadDeSevillaEsBasura is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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3 Answers
3
active
oldest
votes
$begingroup$
Assuming I'm understanding the problem statement correctly and we have a vertical load transferred to a beam via a sort of angled frame, then here's how I'd think about the beam reactions.
The Quick Way...
The quickest approach is to solve for the support reactions by remembering the structure must be globally stable. In this case, we can consider the beam and the frame as a single structure for purposes of calculating the support reactions. Our equations of statics say the sum of the forces in the horizontal direction, the sum of the force in the vertical direction, and sum of the moments, must each be zero. Because there is no externally-applied horizontal load, there is only one possible horizontal force (the support reaction). For the sum of horizontal forces to be zero, the horizontal support reaction must therefore also be zero.
The Longer Way...
The longer way round is perhaps to convince ourselves that the frame applies equal and opposite horizontal loads to the beam and that this, again, means the horizontal support reaction must be zero.
First, imagine we have a simply supported beam with equal and opposite horizontal load applied at a point.
There is no net load at the point, so it's reasonably easy to convince ourselves that there's no support reaction.
What if we applied equal and opposite loads at separate points along the beam?
Now, it's perhaps less clear that there's no support reaction. But we can convince ourselves easily using the equations of statics. The beam reactions are about global stability of the structure. As is always the case in statics, the sum of horizontal forces must be zero.
$$Sigma F_x = (+P) + (-P) + H_A = 0$$
So, by some quick math, we prove that the horizontal reaction at Point A must be zero. In fact, for this beam configuration, the only segment of the beam "feeling" axial load is the segment in between the applied loads. It can be tempted to picture the left load "pushing" on the beam segment to the left of it, but that's not what's happening. Imagine a game of tug-of-war where the two innermost opponents perfectly oppose each other's forces...the players farther down the rope wouldn't feel anything, right?
So now, all we have to do is convince ourselves the horizontal components of the applied loads are equal and opposite.
If we zoomed in and just looked at the bit of structure right were the vertical load is applied...well, that bit of structure must also be in equilibrium. The applied load is going to distribute between the two angled members, which means there will be horizontal components. However, again, by our lovely equations of statics, for this bit of structure to be in equilibrium, the sum of horizontal forces must be zero. Here, the only possible horizontal forces are the horizontal components of the loads in the angled members -- and thus, these components must be equal and opposite.
Ta, da! Sometimes our initial intuition can lead us a little astray but the equations of statics never lie. Forces and moments must always sum to zero in order for the structure to remain...static.
answered Jun 16 at 18:36
CableStayCableStay
1,8121 gold badge8 silver badges25 bronze badges
$endgroup$
$begingroup$
Thank you, for this clear information. Very appreciated.
$endgroup$
– UniversidadDeSevillaEsBasura
Jun 16 at 18:51
add a comment |
$begingroup$
Think about the three members that form the triangle EBC.
There is no horizontal load applied at E.
You know there is no horizontal load applied at C by beam CD, because there is a roller bearing at D.
Therefore by equilibrium, there is no horizontal load applied at B.
Therefore there is no horizontal tension in beam AB and no horizontal reaction at A.
Of course there will be horizontal tension in BC, but that will be balanced at each end by the horizontal components of tension in EB and EC.
answered Jun 16 at 14:51
alephzeroalephzero
8,0481 gold badge6 silver badges22 bronze badges
$endgroup$
$begingroup$
Thanks for your time. But that is my confusion. I don't want to look at E and C where nothing is really happening. I want to look at A and B, where B is actually "pushing" against A, and so A should push back. Is there any way of visualizing this, only looking at A and B, where the fancy stuff is actually happening? Thanks a lot for your time, again.
$endgroup$
– UniversidadDeSevillaEsBasura
Jun 16 at 17:22
$begingroup$
You seem to be asking "can I solve the problem by guessing which features I think are interesting and ignoring the rest? The answer is obviously "no, because you are not clever enough to guess right."
$endgroup$
– alephzero
Jun 16 at 19:10
add a comment |
$begingroup$
You are more than overthinking this. Keep it simple. For an object at rest:
- ΣF(x) = 0 [sum of vertical forces = zero]
- ΣF(y) = 0 [sum of horizontal forces = zero]
ΣMoments() = 0 [sum of moments about any point = zero]
Free body diagram of beam.
Again, ΣF(x) = 0. With the diagram drawn this way, what do you thing Force A (in the x direction) is?
answered Jun 17 at 1:09
zipzitzipzit
2012 silver badges7 bronze badges
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add a comment |
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3 Answers
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3 Answers
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$begingroup$
Assuming I'm understanding the problem statement correctly and we have a vertical load transferred to a beam via a sort of angled frame, then here's how I'd think about the beam reactions.
The Quick Way...
The quickest approach is to solve for the support reactions by remembering the structure must be globally stable. In this case, we can consider the beam and the frame as a single structure for purposes of calculating the support reactions. Our equations of statics say the sum of the forces in the horizontal direction, the sum of the force in the vertical direction, and sum of the moments, must each be zero. Because there is no externally-applied horizontal load, there is only one possible horizontal force (the support reaction). For the sum of horizontal forces to be zero, the horizontal support reaction must therefore also be zero.
The Longer Way...
The longer way round is perhaps to convince ourselves that the frame applies equal and opposite horizontal loads to the beam and that this, again, means the horizontal support reaction must be zero.
First, imagine we have a simply supported beam with equal and opposite horizontal load applied at a point.
There is no net load at the point, so it's reasonably easy to convince ourselves that there's no support reaction.
What if we applied equal and opposite loads at separate points along the beam?
Now, it's perhaps less clear that there's no support reaction. But we can convince ourselves easily using the equations of statics. The beam reactions are about global stability of the structure. As is always the case in statics, the sum of horizontal forces must be zero.
$$Sigma F_x = (+P) + (-P) + H_A = 0$$
So, by some quick math, we prove that the horizontal reaction at Point A must be zero. In fact, for this beam configuration, the only segment of the beam "feeling" axial load is the segment in between the applied loads. It can be tempted to picture the left load "pushing" on the beam segment to the left of it, but that's not what's happening. Imagine a game of tug-of-war where the two innermost opponents perfectly oppose each other's forces...the players farther down the rope wouldn't feel anything, right?
So now, all we have to do is convince ourselves the horizontal components of the applied loads are equal and opposite.
If we zoomed in and just looked at the bit of structure right were the vertical load is applied...well, that bit of structure must also be in equilibrium. The applied load is going to distribute between the two angled members, which means there will be horizontal components. However, again, by our lovely equations of statics, for this bit of structure to be in equilibrium, the sum of horizontal forces must be zero. Here, the only possible horizontal forces are the horizontal components of the loads in the angled members -- and thus, these components must be equal and opposite.
Ta, da! Sometimes our initial intuition can lead us a little astray but the equations of statics never lie. Forces and moments must always sum to zero in order for the structure to remain...static.
answered Jun 16 at 18:36
CableStayCableStay
1,8121 gold badge8 silver badges25 bronze badges
$endgroup$
$begingroup$
Thank you, for this clear information. Very appreciated.
$endgroup$
– UniversidadDeSevillaEsBasura
Jun 16 at 18:51
add a comment |
$begingroup$
Assuming I'm understanding the problem statement correctly and we have a vertical load transferred to a beam via a sort of angled frame, then here's how I'd think about the beam reactions.
The Quick Way...
The quickest approach is to solve for the support reactions by remembering the structure must be globally stable. In this case, we can consider the beam and the frame as a single structure for purposes of calculating the support reactions. Our equations of statics say the sum of the forces in the horizontal direction, the sum of the force in the vertical direction, and sum of the moments, must each be zero. Because there is no externally-applied horizontal load, there is only one possible horizontal force (the support reaction). For the sum of horizontal forces to be zero, the horizontal support reaction must therefore also be zero.
The Longer Way...
The longer way round is perhaps to convince ourselves that the frame applies equal and opposite horizontal loads to the beam and that this, again, means the horizontal support reaction must be zero.
First, imagine we have a simply supported beam with equal and opposite horizontal load applied at a point.
There is no net load at the point, so it's reasonably easy to convince ourselves that there's no support reaction.
What if we applied equal and opposite loads at separate points along the beam?
Now, it's perhaps less clear that there's no support reaction. But we can convince ourselves easily using the equations of statics. The beam reactions are about global stability of the structure. As is always the case in statics, the sum of horizontal forces must be zero.
$$Sigma F_x = (+P) + (-P) + H_A = 0$$
So, by some quick math, we prove that the horizontal reaction at Point A must be zero. In fact, for this beam configuration, the only segment of the beam "feeling" axial load is the segment in between the applied loads. It can be tempted to picture the left load "pushing" on the beam segment to the left of it, but that's not what's happening. Imagine a game of tug-of-war where the two innermost opponents perfectly oppose each other's forces...the players farther down the rope wouldn't feel anything, right?
So now, all we have to do is convince ourselves the horizontal components of the applied loads are equal and opposite.
If we zoomed in and just looked at the bit of structure right were the vertical load is applied...well, that bit of structure must also be in equilibrium. The applied load is going to distribute between the two angled members, which means there will be horizontal components. However, again, by our lovely equations of statics, for this bit of structure to be in equilibrium, the sum of horizontal forces must be zero. Here, the only possible horizontal forces are the horizontal components of the loads in the angled members -- and thus, these components must be equal and opposite.
Ta, da! Sometimes our initial intuition can lead us a little astray but the equations of statics never lie. Forces and moments must always sum to zero in order for the structure to remain...static.
answered Jun 16 at 18:36
CableStayCableStay
1,8121 gold badge8 silver badges25 bronze badges
$endgroup$
$begingroup$
Thank you, for this clear information. Very appreciated.
$endgroup$
– UniversidadDeSevillaEsBasura
Jun 16 at 18:51
add a comment |
$begingroup$
Assuming I'm understanding the problem statement correctly and we have a vertical load transferred to a beam via a sort of angled frame, then here's how I'd think about the beam reactions.
The Quick Way...
The quickest approach is to solve for the support reactions by remembering the structure must be globally stable. In this case, we can consider the beam and the frame as a single structure for purposes of calculating the support reactions. Our equations of statics say the sum of the forces in the horizontal direction, the sum of the force in the vertical direction, and sum of the moments, must each be zero. Because there is no externally-applied horizontal load, there is only one possible horizontal force (the support reaction). For the sum of horizontal forces to be zero, the horizontal support reaction must therefore also be zero.
The Longer Way...
The longer way round is perhaps to convince ourselves that the frame applies equal and opposite horizontal loads to the beam and that this, again, means the horizontal support reaction must be zero.
First, imagine we have a simply supported beam with equal and opposite horizontal load applied at a point.
There is no net load at the point, so it's reasonably easy to convince ourselves that there's no support reaction.
What if we applied equal and opposite loads at separate points along the beam?
Now, it's perhaps less clear that there's no support reaction. But we can convince ourselves easily using the equations of statics. The beam reactions are about global stability of the structure. As is always the case in statics, the sum of horizontal forces must be zero.
$$Sigma F_x = (+P) + (-P) + H_A = 0$$
So, by some quick math, we prove that the horizontal reaction at Point A must be zero. In fact, for this beam configuration, the only segment of the beam "feeling" axial load is the segment in between the applied loads. It can be tempted to picture the left load "pushing" on the beam segment to the left of it, but that's not what's happening. Imagine a game of tug-of-war where the two innermost opponents perfectly oppose each other's forces...the players farther down the rope wouldn't feel anything, right?
So now, all we have to do is convince ourselves the horizontal components of the applied loads are equal and opposite.
If we zoomed in and just looked at the bit of structure right were the vertical load is applied...well, that bit of structure must also be in equilibrium. The applied load is going to distribute between the two angled members, which means there will be horizontal components. However, again, by our lovely equations of statics, for this bit of structure to be in equilibrium, the sum of horizontal forces must be zero. Here, the only possible horizontal forces are the horizontal components of the loads in the angled members -- and thus, these components must be equal and opposite.
Ta, da! Sometimes our initial intuition can lead us a little astray but the equations of statics never lie. Forces and moments must always sum to zero in order for the structure to remain...static.
answered Jun 16 at 18:36
CableStayCableStay
1,8121 gold badge8 silver badges25 bronze badges
$endgroup$
Assuming I'm understanding the problem statement correctly and we have a vertical load transferred to a beam via a sort of angled frame, then here's how I'd think about the beam reactions.
The Quick Way...
The quickest approach is to solve for the support reactions by remembering the structure must be globally stable. In this case, we can consider the beam and the frame as a single structure for purposes of calculating the support reactions. Our equations of statics say the sum of the forces in the horizontal direction, the sum of the force in the vertical direction, and sum of the moments, must each be zero. Because there is no externally-applied horizontal load, there is only one possible horizontal force (the support reaction). For the sum of horizontal forces to be zero, the horizontal support reaction must therefore also be zero.
The Longer Way...
The longer way round is perhaps to convince ourselves that the frame applies equal and opposite horizontal loads to the beam and that this, again, means the horizontal support reaction must be zero.
First, imagine we have a simply supported beam with equal and opposite horizontal load applied at a point.
There is no net load at the point, so it's reasonably easy to convince ourselves that there's no support reaction.
What if we applied equal and opposite loads at separate points along the beam?
Now, it's perhaps less clear that there's no support reaction. But we can convince ourselves easily using the equations of statics. The beam reactions are about global stability of the structure. As is always the case in statics, the sum of horizontal forces must be zero.
$$Sigma F_x = (+P) + (-P) + H_A = 0$$
So, by some quick math, we prove that the horizontal reaction at Point A must be zero. In fact, for this beam configuration, the only segment of the beam "feeling" axial load is the segment in between the applied loads. It can be tempted to picture the left load "pushing" on the beam segment to the left of it, but that's not what's happening. Imagine a game of tug-of-war where the two innermost opponents perfectly oppose each other's forces...the players farther down the rope wouldn't feel anything, right?
So now, all we have to do is convince ourselves the horizontal components of the applied loads are equal and opposite.
If we zoomed in and just looked at the bit of structure right were the vertical load is applied...well, that bit of structure must also be in equilibrium. The applied load is going to distribute between the two angled members, which means there will be horizontal components. However, again, by our lovely equations of statics, for this bit of structure to be in equilibrium, the sum of horizontal forces must be zero. Here, the only possible horizontal forces are the horizontal components of the loads in the angled members -- and thus, these components must be equal and opposite.
Ta, da! Sometimes our initial intuition can lead us a little astray but the equations of statics never lie. Forces and moments must always sum to zero in order for the structure to remain...static.
answered Jun 16 at 18:36
CableStayCableStay
1,8121 gold badge8 silver badges25 bronze badges
answered Jun 16 at 18:36
CableStayCableStay
1,8121 gold badge8 silver badges25 bronze badges
answered Jun 16 at 18:36
CableStayCableStay
1,8121 gold badge8 silver badges25 bronze badges
answered Jun 16 at 18:36
CableStayCableStay
1,8121 gold badge8 silver badges25 bronze badges
1,8121 gold badge8 silver badges25 bronze badges
$begingroup$
Thank you, for this clear information. Very appreciated.
$endgroup$
– UniversidadDeSevillaEsBasura
Jun 16 at 18:51
add a comment |
$begingroup$
Thank you, for this clear information. Very appreciated.
$endgroup$
– UniversidadDeSevillaEsBasura
Jun 16 at 18:51
$begingroup$
Thank you, for this clear information. Very appreciated.
$endgroup$
– UniversidadDeSevillaEsBasura
Jun 16 at 18:51
$begingroup$
Thank you, for this clear information. Very appreciated.
$endgroup$
– UniversidadDeSevillaEsBasura
Jun 16 at 18:51
add a comment |
$begingroup$
Think about the three members that form the triangle EBC.
There is no horizontal load applied at E.
You know there is no horizontal load applied at C by beam CD, because there is a roller bearing at D.
Therefore by equilibrium, there is no horizontal load applied at B.
Therefore there is no horizontal tension in beam AB and no horizontal reaction at A.
Of course there will be horizontal tension in BC, but that will be balanced at each end by the horizontal components of tension in EB and EC.
answered Jun 16 at 14:51
alephzeroalephzero
8,0481 gold badge6 silver badges22 bronze badges
$endgroup$
$begingroup$
Thanks for your time. But that is my confusion. I don't want to look at E and C where nothing is really happening. I want to look at A and B, where B is actually "pushing" against A, and so A should push back. Is there any way of visualizing this, only looking at A and B, where the fancy stuff is actually happening? Thanks a lot for your time, again.
$endgroup$
– UniversidadDeSevillaEsBasura
Jun 16 at 17:22
$begingroup$
You seem to be asking "can I solve the problem by guessing which features I think are interesting and ignoring the rest? The answer is obviously "no, because you are not clever enough to guess right."
$endgroup$
– alephzero
Jun 16 at 19:10
add a comment |
$begingroup$
Think about the three members that form the triangle EBC.
There is no horizontal load applied at E.
You know there is no horizontal load applied at C by beam CD, because there is a roller bearing at D.
Therefore by equilibrium, there is no horizontal load applied at B.
Therefore there is no horizontal tension in beam AB and no horizontal reaction at A.
Of course there will be horizontal tension in BC, but that will be balanced at each end by the horizontal components of tension in EB and EC.
answered Jun 16 at 14:51
alephzeroalephzero
8,0481 gold badge6 silver badges22 bronze badges
$endgroup$
$begingroup$
Thanks for your time. But that is my confusion. I don't want to look at E and C where nothing is really happening. I want to look at A and B, where B is actually "pushing" against A, and so A should push back. Is there any way of visualizing this, only looking at A and B, where the fancy stuff is actually happening? Thanks a lot for your time, again.
$endgroup$
– UniversidadDeSevillaEsBasura
Jun 16 at 17:22
$begingroup$
You seem to be asking "can I solve the problem by guessing which features I think are interesting and ignoring the rest? The answer is obviously "no, because you are not clever enough to guess right."
$endgroup$
– alephzero
Jun 16 at 19:10
add a comment |
$begingroup$
Think about the three members that form the triangle EBC.
There is no horizontal load applied at E.
You know there is no horizontal load applied at C by beam CD, because there is a roller bearing at D.
Therefore by equilibrium, there is no horizontal load applied at B.
Therefore there is no horizontal tension in beam AB and no horizontal reaction at A.
Of course there will be horizontal tension in BC, but that will be balanced at each end by the horizontal components of tension in EB and EC.
answered Jun 16 at 14:51
alephzeroalephzero
8,0481 gold badge6 silver badges22 bronze badges
$endgroup$
Think about the three members that form the triangle EBC.
There is no horizontal load applied at E.
You know there is no horizontal load applied at C by beam CD, because there is a roller bearing at D.
Therefore by equilibrium, there is no horizontal load applied at B.
Therefore there is no horizontal tension in beam AB and no horizontal reaction at A.
Of course there will be horizontal tension in BC, but that will be balanced at each end by the horizontal components of tension in EB and EC.
answered Jun 16 at 14:51
alephzeroalephzero
8,0481 gold badge6 silver badges22 bronze badges
answered Jun 16 at 14:51
alephzeroalephzero
8,0481 gold badge6 silver badges22 bronze badges
answered Jun 16 at 14:51
alephzeroalephzero
8,0481 gold badge6 silver badges22 bronze badges
answered Jun 16 at 14:51
alephzeroalephzero
8,0481 gold badge6 silver badges22 bronze badges
8,0481 gold badge6 silver badges22 bronze badges
$begingroup$
Thanks for your time. But that is my confusion. I don't want to look at E and C where nothing is really happening. I want to look at A and B, where B is actually "pushing" against A, and so A should push back. Is there any way of visualizing this, only looking at A and B, where the fancy stuff is actually happening? Thanks a lot for your time, again.
$endgroup$
– UniversidadDeSevillaEsBasura
Jun 16 at 17:22
$begingroup$
You seem to be asking "can I solve the problem by guessing which features I think are interesting and ignoring the rest? The answer is obviously "no, because you are not clever enough to guess right."
$endgroup$
– alephzero
Jun 16 at 19:10
add a comment |
$begingroup$
Thanks for your time. But that is my confusion. I don't want to look at E and C where nothing is really happening. I want to look at A and B, where B is actually "pushing" against A, and so A should push back. Is there any way of visualizing this, only looking at A and B, where the fancy stuff is actually happening? Thanks a lot for your time, again.
$endgroup$
– UniversidadDeSevillaEsBasura
Jun 16 at 17:22
$begingroup$
You seem to be asking "can I solve the problem by guessing which features I think are interesting and ignoring the rest? The answer is obviously "no, because you are not clever enough to guess right."
$endgroup$
– alephzero
Jun 16 at 19:10
$begingroup$
Thanks for your time. But that is my confusion. I don't want to look at E and C where nothing is really happening. I want to look at A and B, where B is actually "pushing" against A, and so A should push back. Is there any way of visualizing this, only looking at A and B, where the fancy stuff is actually happening? Thanks a lot for your time, again.
$endgroup$
– UniversidadDeSevillaEsBasura
Jun 16 at 17:22
$begingroup$
Thanks for your time. But that is my confusion. I don't want to look at E and C where nothing is really happening. I want to look at A and B, where B is actually "pushing" against A, and so A should push back. Is there any way of visualizing this, only looking at A and B, where the fancy stuff is actually happening? Thanks a lot for your time, again.
$endgroup$
– UniversidadDeSevillaEsBasura
Jun 16 at 17:22
$begingroup$
You seem to be asking "can I solve the problem by guessing which features I think are interesting and ignoring the rest? The answer is obviously "no, because you are not clever enough to guess right."
$endgroup$
– alephzero
Jun 16 at 19:10
$begingroup$
You seem to be asking "can I solve the problem by guessing which features I think are interesting and ignoring the rest? The answer is obviously "no, because you are not clever enough to guess right."
$endgroup$
– alephzero
Jun 16 at 19:10
add a comment |
$begingroup$
You are more than overthinking this. Keep it simple. For an object at rest:
- ΣF(x) = 0 [sum of vertical forces = zero]
- ΣF(y) = 0 [sum of horizontal forces = zero]
ΣMoments() = 0 [sum of moments about any point = zero]
Free body diagram of beam.
Again, ΣF(x) = 0. With the diagram drawn this way, what do you thing Force A (in the x direction) is?
answered Jun 17 at 1:09
zipzitzipzit
2012 silver badges7 bronze badges
$endgroup$
add a comment |
$begingroup$
You are more than overthinking this. Keep it simple. For an object at rest:
- ΣF(x) = 0 [sum of vertical forces = zero]
- ΣF(y) = 0 [sum of horizontal forces = zero]
ΣMoments() = 0 [sum of moments about any point = zero]
Free body diagram of beam.
Again, ΣF(x) = 0. With the diagram drawn this way, what do you thing Force A (in the x direction) is?
answered Jun 17 at 1:09
zipzitzipzit
2012 silver badges7 bronze badges
$endgroup$
add a comment |
$begingroup$
You are more than overthinking this. Keep it simple. For an object at rest:
- ΣF(x) = 0 [sum of vertical forces = zero]
- ΣF(y) = 0 [sum of horizontal forces = zero]
ΣMoments() = 0 [sum of moments about any point = zero]
Free body diagram of beam.
Again, ΣF(x) = 0. With the diagram drawn this way, what do you thing Force A (in the x direction) is?
answered Jun 17 at 1:09
zipzitzipzit
2012 silver badges7 bronze badges
$endgroup$
You are more than overthinking this. Keep it simple. For an object at rest:
- ΣF(x) = 0 [sum of vertical forces = zero]
- ΣF(y) = 0 [sum of horizontal forces = zero]
ΣMoments() = 0 [sum of moments about any point = zero]
Free body diagram of beam.
Again, ΣF(x) = 0. With the diagram drawn this way, what do you thing Force A (in the x direction) is?
answered Jun 17 at 1:09
zipzitzipzit
2012 silver badges7 bronze badges
answered Jun 17 at 1:09
zipzitzipzit
2012 silver badges7 bronze badges
answered Jun 17 at 1:09
zipzitzipzit
2012 silver badges7 bronze badges
answered Jun 17 at 1:09
zipzitzipzit
2012 silver badges7 bronze badges
2012 silver badges7 bronze badges
add a comment |
add a comment |
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