bending moment, shear and normal diagrams
DESCRIPTION
This presentation is part of the Solid Mechanics course taught in the second year of the Degree in Architecture at San Pablo CEU University Institute of Technology, in Madrid (Spain). Three step by step examples of how to draw bending moment, shear and normal force diagrams in 2D, statically determinate structures, are depicted. Feel free to comment!TRANSCRIPT
Solid Mechanics Academic Year 2014/2015
Solid Mechanics
Block A. Internal forces diagrams.
Prof. Maribel Castilla Heredia @maribelcastilla Version 1.0 October 2014
Degree in Architecture.
San Pablo CEU University – Institute of Technology.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces
As seen during the classes, we’ll think of our structural members as if they consisted
of thin slices along the axis of the bar. When external forces are applied to our
structural system, the slices are subjected to internal forces.
These internal forces exerted on the slices appear in couples, as each of the slices
must be in equilibrium itself.
For a plane structural member, there are three possible kinds of internal forces:
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Axial (or Normal) force
It makes both sides of the slice
separate one from another, keeping
both sides parallel.
Letter N represents axial force.
Internal forces
As seen during the classes, we’ll think of our structural members as if they consisted
of thin slices along the axis of the bar. When external forces are applied to our
structural system, the slices are subjected to internal forces.
These internal forces exerted on the slices appear in couples, as each of the slices
must be in equilibrium itself.
For a plane structural member, there are three possible kinds of internal forces:
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces
As seen during the classes, we’ll think of our structural members as if they consisted
of thin slices along the axis of the bar. When external forces are applied to our
structural system, the slices are subjected to internal forces.
These internal forces exerted on the slices appear in couples, as each of the slices
must be in equilibrium itself.
For a plane structural member, there are three possible kinds of internal forces:
Shear force
It makes both sides of the slice slide
one from another perpendicularly to
the axis of the bar. Both sides keep
parallel.
Letter V represents shear.
Axial (or Normal) force
It makes both sides of the slice
separate one from another, keeping
both sides parallel.
Letter N represents axial force.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces
As seen during the classes, we’ll think of our structural members as if they consisted
of thin slices along the axis of the bar. When external forces are applied to our
structural system, the slices are subjected to internal forces.
These internal forces exerted on the slices appear in couples, as each of the slices
must be in equilibrium itself.
For a plane structural member, there are three possible kinds of internal forces:
Shear force
It makes both sides of the slice slide
one from another perpendicularly to
the axis of the bar. Both sides keep
parallel.
Letter V represents shear.
Axial (or Normal) force
It makes both sides of the slice
separate one from another, keeping
both sides parallel.
Letter N represents axial force.
Bending moment
It makes both sides of the slice rotate,
so they stop being parallel. As for the
upper and lower sides of the slice, one
shortens and the other lengthens.
Letter M represents bending moment.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces
As seen during the classes, we’ll think of our structural members as if they consisted
of thin slices along the axis of the bar. When external forces are applied to our
structural system, the slices are subjected to internal forces.
These internal forces exerted on the slices appear in couples, as each of the slices
must be in equilibrium itself.
For a plane structural member, there are three possible kinds of internal forces:
Shear force
It makes both sides of the slice slide
one from another perpendicularly to
the axis of the bar. Both sides keep
parallel.
Letter V represents shear.
Axial (or Normal) force
It makes both sides of the slice
separate one from another, keeping
both sides parallel.
Letter N represents axial force.
Bending moment
It makes both sides of the slice rotate,
so they stop being parallel. As for the
upper and lower sides of the slice, one
shortens and the other lengthens.
Letter M represents bending moment.
Check this out!
If these slices are equilibrated, both acting forces have to be equal in magnitude but
opposed in sense.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces. Sign criteria.
Internal forces are always defined by couples of forces (or moments) that act on a slice that
belongs to a structural member. Thus, the sign can’t be stablished just buy looking at the sense of
one of these forces, but at the global effect that those two forces produce on the slice.
Sign criteria are arbitrary. Before analyzing any structural system we must be aware of the sign
criteria we are expected to use. Likewise, if I want to read the results of the analysis of a structural
system that someone (other than me) has calculated, I must know first which sign criteria this
person has employed.
The following criteria is the one that we use throughout the course, but it doesn’t have to be the
same as the one used in other courses, textbooks, etc.
This examples are valid for members with horizontal axis (horizontal bars).
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces. Sign criteria.
Internal forces are always defined by couples of forces (or moments) that act on a slice that
belongs to a structural member. Thus, the sign can’t be stablished just buy looking at the sense of
one of these forces, but at the global effect that those two forces produce on the slice.
Sign criteria are arbitrary. Before analyzing any structural system we must be aware of the sign
criteria we are expected to use. Likewise, if I want to read the results of the analysis of a structural
system that someone (other than me) has calculated, I must know first which sign criteria this
person has employed.
The following criteria is the one that we use throughout the course, but it doesn’t have to be the
same as the one used in other courses, textbooks, etc.
This examples are valid for members with horizontal axis (horizontal bars).
Axial force
If both sides of the slice tend to separate one
from another, the sign will be positive. If they
tend to get closer, the sign will be negative.
When axial force is positive we say that the
member is subjected to tension. If negative,
we say the member is compressed.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces. Sign criteria.
Internal forces are always defined by couples of forces (or moments) that act on a slice that
belongs to a structural member. Thus, the sign can’t be stablished just buy looking at the sense of
one of these forces, but at the global effect that those two forces produce on the slice.
Sign criteria are arbitrary. Before analyzing any structural system we must be aware of the sign
criteria we are expected to use. Likewise, if I want to read the results of the analysis of a structural
system that someone (other than me) has calculated, I must know first which sign criteria this
person has employed.
The following criteria is the one that we use throughout the course, but it doesn’t have to be the
same as the one used in other courses, textbooks, etc.
This examples are valid for members with horizontal axis (horizontal bars).
Axial force
If both sides of the slice tend to separate one
from another, the sign will be positive. If they
tend to get closer, the sign will be negative.
When axial force is positive we say that the
member is subjected to tension. If negative,
we say the member is compressed.
Shear force
We will say shear force is positive if the
rotation that both forces would produce on
the slice is a counterclockwise one and
negative if the rotation were a clockwise one.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces. Sign criteria.
Internal forces are always defined by couples of forces (or moments) that act on a slice that
belongs to a structural member. Thus, the sign can’t be stablished just buy looking at the sense of
one of these forces, but at the global effect that those two forces produce on the slice.
Sign criteria are arbitrary. Before analyzing any structural system we must be aware of the sign
criteria we are expected to use. Likewise, if I want to read the results of the analysis of a structural
system that someone (other than me) has calculated, I must know first which sign criteria this
person has employed.
The following criteria is the one that we use throughout the course, but it doesn’t have to be the
same as the one used in other courses, textbooks, etc.
This examples are valid for members with horizontal axis (horizontal bars).
Axial force
If both sides of the slice tend to separate one
from another, the sign will be positive. If they
tend to get closer, the sign will be negative.
When axial force is positive we say that the
member is subjected to tension. If negative,
we say the member is compressed.
Shear force
We will say shear force is positive if the
rotation that both forces would produce on
the slice is a counterclockwise one and
negative if the rotation were a clockwise one.
Bending moment
It will be considered positive if the lower side
of the slice tends to lengthen. This implies
that the upper side is compressed and the
lower side is subjected to tension..
Values in bending moment diagrams will
always be drawn beside the tensioned side of
the member.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Criterio de signos a emplear en esfuerzos (II)
Sign criteria is linked to the behaviour of slices. The orientation of the slices depend on the
orientation of the member they belong to. That’s why sign criteria must be referred to a local
coordinate system or to a representation of the slices in the different positions they can be found
along the structural system.
The following criteria is the one we will be using throughout the course: the three types of internal
forces are represented in their positive disposition on one single slice. Then, we represent this slice
in each possible position we could find it in the problems that we will have to solve.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Horizontal member
Criterio de signos a emplear en esfuerzos (II)
Sign criteria is linked to the behaviour of slices. The orientation of the slices depend on the
orientation of the member they belong to. That’s why sign criteria must be referred to a local
coordinate system or to a representation of the slices in the different positions they can be found
along the structural system.
The following criteria is the one we will be using throughout the course: the three types of internal
forces are represented in their positive disposition on one single slice. Then, we represent this slice
in each possible position we could find it in the problems that we will have to solve.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Horizontal member
Criterio de signos a emplear en esfuerzos (II)
Sign criteria is linked to the behaviour of slices. The orientation of the slices depend on the
orientation of the member they belong to. That’s why sign criteria must be referred to a local
coordinate system or to a representation of the slices in the different positions they can be found
along the structural system.
The following criteria is the one we will be using throughout the course: the three types of internal
forces are represented in their positive disposition on one single slice. Then, we represent this slice
in each possible position we could find it in the problems that we will have to solve.
Vertical member
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Horizontal member
Criterio de signos a emplear en esfuerzos (II)
Sign criteria is linked to the behaviour of slices. The orientation of the slices depend on the
orientation of the member they belong to. That’s why sign criteria must be referred to a local
coordinate system or to a representation of the slices in the different positions they can be found
along the structural system.
The following criteria is the one we will be using throughout the course: the three types of internal
forces are represented in their positive disposition on one single slice. Then, we represent this slice
in each possible position we could find it in the problems that we will have to solve.
Inclined member. Increasing slope.
Vertical member
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Horizontal member
Criterio de signos a emplear en esfuerzos (II)
Sign criteria is linked to the behaviour of slices. The orientation of the slices depend on the
orientation of the member they belong to. That’s why sign criteria must be referred to a local
coordinate system or to a representation of the slices in the different positions they can be found
along the structural system.
The following criteria is the one we will be using throughout the course: the three types of internal
forces are represented in their positive disposition on one single slice. Then, we represent this slice
in each possible position we could find it in the problems that we will have to solve.
Inclined member. Increasing slope.
Vertical member
Inclined member. Decreasing slope.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Diagramas de esfuerzos
To obtain the values of the internal forces in different points of the structure, we will
“cut” the structure at the desired point. This “cut” will show three unknowns (the
internal forces I can see in the visible side of the last slice before the cut). Using the
equilibrium equations I will find the value of these unknowns, will give me the values
for the axial and shear forces and the bending moment at that point.
We must learn how axial, shear and bending moment vary along each of the members
of the structure, so that we become able to interpret the diagrams with just a glimpse.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Diagramas de esfuerzos
To obtain the values of the internal forces in different points of the structure, we will
“cut” the structure at the desired point. This “cut” will show three unknowns (the
internal forces I can see in the visible side of the last slice before the cut). Using the
equilibrium equations I will find the value of these unknowns, will give me the values
for the axial and shear forces and the bending moment at that point.
We must learn how axial, shear and bending moment vary along each of the members
of the structure, so that we become able to interpret the diagrams with just a glimpse.
Overall procedure:
Obtain the reactions at the supports (or internal joints) and make sure that the structural system has
been correctly equilibrated.
Cut the structure at the needed points and obtain the values for the internal forces at each of those
points. Depict these values on the correct side of the structural member.
Represent the functions that will eventually form our diagrams.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. First example.
Let’s limber up with the simply supported beam.
We apply a point load on it and obtain the reactions.
We will obtain the values of the internal forces in several sections of the bar and we will
deduce which are the actual points at which we need to obtain these values.
Finally, we will draw the diagram for each internal force.
As we want to know how internal forces vary along the member, we must stablish an “x” axis
and a local coordinate system.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. First example.
Let’s limber up with the simply supported beam.
We apply a point load on it and obtain the reactions.
We will obtain the values of the internal forces in several sections of the bar and we will
deduce which are the actual points at which we need to obtain these values.
Finally, we will draw the diagram for each internal force.
As we want to know how internal forces vary along the member, we must stablish an “x” axis
and a local coordinate system.
The axis of the bar will be the “x” axis and the origin will be located at the left support..
The red point represents the point at which I’ll be cutting the structure.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. First example (cont).
Let’s obtain the values for the internal forces at different points of the member.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. First example (cont).
Let’s obtain the values for the internal forces at different points of the member.
X = 0
- I cut the structure in two infinitely near the left support and I keep the smallest
part (this would be the last slice before the support). Once I cut, I can see the
forces on the right side of the slice, which will be represented according to the
positive sign criteria for a horizontal bar.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. First example (cont).
Let’s obtain the values for the internal forces at different points of the member.
X = 0
- I cut the structure in two infinitely near the left support and I keep the smallest
part (this would be the last slice before the support). Once I cut, I can see the
forces on the right side of the slice, which will be represented according to the
positive sign criteria for a horizontal bar.
- I equilibrate the remaining forces and moments on the structure according to the
direction of each of my unknowns:
- Axial force: There are no forces in the direction of the axial force for “N” to
equilibrate, so the value will be zero.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. First example (cont).
Let’s obtain the values for the internal forces at different points of the member.
X = 0
- I cut the structure in two infinitely near the left support and I keep the smallest
part (this would be the last slice before the support). Once I cut, I can see the
forces on the right side of the slice, which will be represented according to the
positive sign criteria for a horizontal bar.
- I equilibrate the remaining forces and moments on the structure according to the
direction of each of my unknowns:
- Axial force: There are no forces in the direction of the axial force for “N” to
equilibrate, so the value will be zero.
- Shear: The only acting force in the shear direction (perpendicular to the axis of
the member) is the vertical reaction. Therefore, the magnitude of the shear at the
selected point of the member is P/2. We still have to decide on the sign. If we
keep “V” in the sense it would have if it were positive, it’s easy to see that it
wouldn’t be able to equilibrate the left reaction. This means that the sense of the
shear force at that side of the slice must be opposite to the one we had supposed.
Thus, the shear force is negative.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. First example (cont).
Let’s obtain the values for the internal forces at different points of the member.
X = 0
- I cut the structure in two infinitely near the left support and I keep the smallest
part (this would be the last slice before the support). Once I cut, I can see the
forces on the right side of the slice, which will be represented according to the
positive sign criteria for a horizontal bar.
- I equilibrate the remaining forces and moments on the structure according to the
direction of each of my unknowns:
- Axial force: There are no forces in the direction of the axial force for “N” to
equilibrate, so the value will be zero.
- Shear: The only acting force in the shear direction (perpendicular to the axis of
the member) is the vertical reaction. Therefore, the magnitude of the shear at the
selected point of the member is P/2. We still have to decide on the sign. If we
keep “V” in the sense it would have if it were positive, it’s easy to see that it
wouldn’t be able to equilibrate the left reaction. This means that the sense of the
shear force at that side of the slice must be opposite to the one we had supposed.
Thus, the shear force is negative.
- Bending moment. We perform sum of moments at the point we have cut. All
acting forces pass through the selected point, so the bending moment at that point
has to be zero.
- Finally, I represent on each diagram the obtained values.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. First example (cont).
Let’s obtain the values for the internal forces at different points of the member.
X = 0
- I cut the structure in two infinitely near the left support and I keep the smallest
part (this would be the last slice before the support). Once I cut, I can see the
forces on the right side of the slice, which will be represented according to the
positive sign criteria for a horizontal bar.
- I equilibrate the remaining forces and moments on the structure according to the
direction of each of my unknowns:
- Axial force: There are no forces in the direction of the axial force for “N” to
equilibrate, so the value will be zero.
- Shear: The only acting force in the shear direction (perpendicular to the axis of
the member) is the vertical reaction. Therefore, the magnitude of the shear at the
selected point of the member is P/2. We still have to decide on the sign. If we
keep “V” in the sense it would have if it were positive, it’s easy to see that it
wouldn’t be able to equilibrate the left reaction. This means that the sense of the
shear force at that side of the slice must be opposite to the one we had supposed.
Thus, the shear force is negative.
- Bending moment. We perform sum of moments at the point we have cut. All
acting forces pass through the selected point, so the bending moment at that point
has to be zero.
- Finally, I represent on each diagram the obtained values.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. First example (cont).
Let’s obtain the values for the internal forces at different points of the member.
X = 0
- I cut the structure in two infinitely near the left support and I keep the smallest
part (this would be the last slice before the support). Once I cut, I can see the
forces on the right side of the slice, which will be represented according to the
positive sign criteria for a horizontal bar.
- I equilibrate the remaining forces on the structure according to the direction of
each of my unknowns:
- Axial force: There are no forces in the direction of the axial force for “N” to
equilibrate, so the value will be zero.
- Shear: The only acting force in the shear direction (perpendicular to the axis of
the member) is the vertical reaction. Therefore, the magnitude of the shear at the
selected point of the member is P/2. We still have to decide on the sign. If we
keep “V” in the sense it would have if it were positive, it’s easy to see that it
wouldn’t be able to equilibrate the left reaction. This means that the sense of the
shear force at that side of the slice must be opposite to the one we had supposed.
Thus, the shear force is negative.
- Bending moment. We perform sum of moments at the point we have cut. All
acting forces pass through the selected point, so the bending moment at that point
has to be zero.
- Finally, I represent on each diagram the obtained values.
Graphical representation
- We will use a Cartesian system in which “y” axis represents the values of each internal force.
- Positive values will be depicted under “x” axis, whereas negative values will be represented over it.
- Once we had obtained several values, we’ll link them with a curve that represents the variation of the
internal forces along the structure.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. First example (cont).
X = L/4
- I split the bar in two at x=L/4 and I keep the smallest part (the left one).
- I equilibrate the acting forces and moments on the left side of the cut with the
unknown internal forces that appear at the right side of the slice, according to my
positive sign criteria:
- Axial force: There are no forces in the direction of the axial force for “N” to
equilibrate, so the value will be zero.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. First example (cont).
X = L/4
- I split the bar in two at x=L/4 and I keep the smallest part (the left one).
- I equilibrate the acting forces and moments on the left side of the cut with the
unknown internal forces that appear at the right side of the slice, according to my
positive sign criteria:
- Axial force: There are no forces in the direction of the axial force for “N” to
equilibrate, so the value will be zero.
- Shear: The only acting force in the shear direction is the vertical reaction.
Therefore, the magnitude of the shear at the selected point of the member is P/2.
If we keep “V” in the sense it would have if it were positive, it wouldn’t be able to
equilibrate the forces on the left side. Thus, the shear force is negative .
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. First example (cont).
X = L/4
- I split the bar in two at x=L/4 and I keep the smallest part (the left one).
- I equilibrate the acting forces and moments on the left side of the cut with the
unknown internal forces that appear at the right side of the slice, according to my
positive sign criteria:
- Axial force: There are no forces in the direction of the axial force for “N” to
equilibrate, so the value will be zero.
- Shear: The only acting force in the shear direction is the vertical reaction.
Therefore, the magnitude of the shear at the selected point of the member is P/2.
If we keep “V” in the sense it would have if it were positive, it wouldn’t be able to
equilibrate the forces on the left side. Thus, the shear force is negative
-Bending moment: The only acting force that can produce a moment is the vertical
reaction. The value or this moment is PL/8, and its direction, clockwise. To
compensate it a moment with the same sense as the one we supposed at first
would be fine, therefore the bending moment is positive.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. First example (cont).
X = L/4
- I split the bar in two at x=L/4 and I keep the smallest part (the left one).
- I equilibrate the acting forces and moments on the left side of the cut with the
unknown internal forces that appear at the right side of the slice, according to my
positive sign criteria:
- Axial force: There are no forces in the direction of the axial force for “N” to
equilibrate, so the value will be zero.
- Shear: The only acting force in the shear direction is the vertical reaction.
Therefore, the magnitude of the shear at the selected point of the member is P/2.
If we keep “V” in the sense it would have if it were positive, it wouldn’t be able to
equilibrate the forces on the left side. Thus, the shear force is negative
-Bending moment: The only acting force that can produce a moment is the vertical
reaction. The value or this moment is PL/8, and its direction, clockwise. To
compensate it a moment with the same sense as the one we supposed at first
would be fine, therefore the bending moment is positive.
- I represent the obtained values on the diagrams.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. First example (cont).
X = L/4
- I split the bar in two at x=L/4 and I keep the smallest part (the left one).
- I equilibrate the acting forces and moments on the left side of the cut with the
unknown internal forces that appear at the right side of the slice, according to my
positive sign criteria:
- Axial force: There are no forces in the direction of the axial force for “N” to
equilibrate, so the value will be zero.
- Shear: The only acting force in the shear direction is the vertical reaction.
Therefore, the magnitude of the shear at the selected point of the member is P/2.
If we keep “V” in the sense it would have if it were positive, it wouldn’t be able to
equilibrate the forces on the left side. Thus, the shear force is negative
-Bending moment: The only acting force that can produce a moment is the vertical
reaction. The value or this moment is PL/8, and its direction, clockwise. To
compensate it a moment with the same sense as the one we supposed at first
would be fine, therefore the bending moment is positive.
- I represent the obtained values on the diagrams.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. First example (cont).
X = L/2 (before the point load)
- I split the bar in two at x=L/2 and I keep the left part.
- I equilibrate the acting forces and moments on the left side of the cut
with the unknown internal forces that appear at the right side of the slice,
according to my positive sign criteria:
- Axial force: zero.
- Shear: Again, the only acting force in the shear direction is the
vertical reaction. Therefore, the magnitude is P/2 and the sign,
negative.
- Bending moment: Again, the only acting force that can produce
a moment is the vertical reaction. But now it’s farther than before.
The magnitude of this moment is PL/8, and its direction,
clockwise Therefore the bending moment is positive.
- I represent the obtained values on the diagrams.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. First example (cont).
X = L/2 (just after the point load)
- I cut the structure again at x=L/2 but this time just after the point load.
- I equilibrate forces and moments on one side and on the other side
of the slice:
- Axial force: zero.
- Shear: The resultant of forces on the left side is P/2 but
downwards. To equilibrate it, the value of the shear has to be
P/2 as well. The sense of the force I had supposed within my
positive sign criteria is suitable this time. Therefore the shear is
positive in this case.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. First example (cont).
X = L/2 (just after the point load)
- I cut the structure again at x=L/2 but this time just after the point load.
- I equilibrate forces and moments on one side and on the other side
of the slice:
- Axial force: zero.
- Shear: The resultant of forces on the left side is P/2 but
downwards. To equilibrate it, the value of the shear has to be
P/2 as well. The sense of the force I had supposed within my
positive sign criteria is suitable this time. Therefore the shear is
positive in this case.
- Bending moment: As I’m performing sum of moments at the
point I’ve cut, the only force producing moments is the left
reaction. Therefore the value of the bending moment is the
same as I had obtained before: + PL/4.
- Again, I represent the obtained values on my diagrams.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. First example (cont).
X = L/2 (just after the point load)
- I cut the structure again at x=L/2 but this time just after the point load.
- I equilibrate the forces and moments one side and on the other side
of the slice:
- Axial force: zero.
- Shear: The resultant of forces on the left side is P/2 but
downwards. To equilibrate it, the value of the shear has to be
P/2 as well. The sense of the force I had supposed within my
positive sign criteria is suitable this time. Therefore the shear is
positive in this case.
- Bending moment: As I’m performing sum of moments at the
point I’ve cut, the only force producing moments is the left
reaction. Therefore the value of the bending moment is the
same as I had obtained before: + PL/4.
- Again, I represent the obtained values on my diagrams.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. First example (cont).
X = 3L/4
- I cut at x=3L/4 and keep the left side of the system (because I want to see the continuous variation).
- I equilibrate the forces on both sides of the cut:
- Axial force: zero.
- Shear: + P/2 .
- Bending moment: in this case both the reaction and the applied load produce moment about the point I have cut
at. The value for the bending moment is + PL/2
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. First example (cont).
X = 3L/4
- I cut at x=3L/4 and keep the left side of the system (because I want to see the continuous variation).
- I equilibrate the forces on both sides of the cut:
- Axial force: zero.
- Shear: + P/2 .
- Bending moment: in this case both the reaction and the applied load produce moment about the point I have cut
at. The value for the bending moment is + PL/2
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. First example (cont).
X = L (just before the right support)
- I cut the structure at x=L just before the right support (so I’m not taking into account the right reaction)
- I equilibrate the forces on both sides of the cut:
- Axial force: zero. · Shear: + P/2.
- Bending moment: Both the left reaction and the point load keep on producing moment about the point I have cut
at, but distances are longer now. The result for the sum of moments produced by acting forces is zero, so the
value for the bending moment is zero as well.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. First example (cont).
X = L (just before the right support)
- I cut the structure at x=L just before the right support (so I’m not taking into account the right reaction)
- I equilibrate the forces on both sides of the cut:
- Axial force: zero. · Shear: + P/2.
- Bending moment: Both the left reaction and the point load keep on producing moment about the point I have cut
at, but distances are longer now. The result for the sum of moments produced by acting forces is zero, so the
value for the bending moment is zero as well.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. First example (cont).
How to plot the diagrams.
- I join with a segment each obtained point and extract some conclusions.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. First example (cont).
How to plot the diagrams.
- I join with a segment each obtained point and extract some conclusions.
- Shear values change when new forces perpendicular to the bar appear.
- Bending moment varies linearly: the farther I cut from the force that produces the moment, the bigger the produced
moment is. Whenever new forces appear, the slope of the diagram changes.
- Conclusion: I only need to cut the members near the points where point forces are applied.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. First example (cont).
How to plot the diagrams.
- I join with a segment each obtained point and extract some conclusions.
- Shear values change when new forces perpendicular to the bar appear.
- Bending moment varies linearly: the farther I cut from the force that produces the moment, the bigger the produced
moment is. Whenever new forces appear, the slope of the diagram changes.
- Conclusion: I only need to cut the members near the points where point forces are applied.
Don’t miss this!
- The “skips” that appear in the shear diagram correspond to the
value of the applied loads at that points.
- Whenever a point load is applied, a skip is produced in the
shear diagram and a change of slope happens in the bending
moment one.
- Positive values in the bending moment diagram represent areas
of the member in which the lower side of the slice is subjected to
tension.
- Bending moment diagrams remember us of the deflection of the
bar.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. Second example.
Simply supported beam with several loads applied on it.
Obtain reactions.
Obtain the values of the internal forces before and after each applied load.
Plot the axial, shear and bending moment diagrams.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. Second example (cont).
Let’s obtain the values for the internal forces at different points of the member.
X = 0
- I cut the structure in two infinitely near the left support and I keep the smallest part
(this would be the last slice before the support). Once I cut, I can see the forces on the
right side of the slice, which will be represented according to the positive sign criteria
for a horizontal bar.
- I equilibrate the remaining forces and moments on the structure according to the
direction of each of my unknowns:
- Axial force: There are no forces in the direction of the axial force for “N” to
equilibrate, so the value will be zero.
- Shear: The only acting force in the shear direction (perpendicular to the axis of the
member) is the vertical reaction. Therefore, the magnitude of the shear at the selected
point of the member is P/2. We still have to decide on the sign. If we keep “V” in the
sense it would have if it were positive, it’s easy to see that it wouldn’t be able to
equilibrate the left reaction. This means that the sense of the shear force at that side of
the slice must be opposite to the one we had supposed. Thus, the shear force is
negative.
- Bending moment. We perform sum of moments at the point we have cut. All acting
forces pass through the selected point, so the bending moment at that point has to be
zero.
- Finally, I represent on each diagram the obtained values.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. Second example (cont).
Let’s obtain the values for the internal forces at different points of the member.
X = 0
- I cut the structure in two infinitely near the left support and I keep the smallest part
(this would be the last slice before the support). Once I cut, I can see the forces on the
right side of the slice, which will be represented according to the positive sign criteria
for a horizontal bar.
- I equilibrate the remaining forces and moments on the structure according to the
direction of each of my unknowns:
- Axial force: There are no forces in the direction of the axial force for “N” to
equilibrate, so the value will be zero.
- Shear: The only acting force in the shear direction (perpendicular to the axis of the
member) is the vertical reaction. Therefore, the magnitude of the shear at the selected
point of the member is P/2. We still have to decide on the sign. If we keep “V” in the
sense it would have if it were positive, it’s easy to see that it wouldn’t be able to
equilibrate the left reaction. This means that the sense of the shear force at that side of
the slice must be opposite to the one we had supposed. Thus, the shear force is
negative.
- Bending moment. We perform sum of moments at the point we have cut. All acting
forces pass through the selected point, so the bending moment at that point has to be
zero.
- Finally, I represent on each diagram the obtained values.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. Second example (cont).
X = L/4 (just before the first applied load)
- I split the bar in two at x=L/4 and I keep the smallest part (the left one).
- I equilibrate the acting forces and moments at the left side of the cut with the
unknown internal forces that appear at the right side of the slice, according to my
positive sign criteria:
- Axial force: There are no forces in the direction of the axial force for “N” to
equilibrate, so the value will be zero.
- Shear: The only acting force in the shear direction is the vertical reaction.
Therefore, the magnitude of the shear at the selected point of the member
is P/2. If we keep “V” in the sense it would have if it were positive, it
wouldn’t be able to equilibrate the forces on the left side. Thus, the shear
force is negative
-Bending moment: The only acting force that can produce a moment is the
vertical reaction. The value or this moment is PL/8, and its direction,
clockwise. To compensate it a moment with the same sense as the one we
supposed at first would be fine, therefore the bending moment is positive.
- I represent the obtained values on the diagrams.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. Second example (cont).
X = L/4 (just after the first point load)
- I split the bar in two at x=L/4 after the point load and keep the left side.
- I equilibrate the acting forces and moments at the left side of the cut with the
unknown internal forces that appear at the right side of the slice, according to my
positive sign criteria:
- Axial force: There are no forces in the direction of the axial force for “N” to
equilibrate, so the value will be zero.
- Shear: The resultant of forces on the left side is P/6 upwards. To
equilibrate it, the value of the shear has to be P/6 as well. The sense of the
force I had supposed within my positive sign criteria is not valid to
compensate this force, so the shear is negative again.
-Bending moment: There are two forces now but only the vertical reaction
can produce a moment about the point I’ve cut at. The value or this
moment is, again, +PL/8.
- I represent the obtained values on the diagrams.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. Second example (cont).
X = L/2 (just before the point load)
- I cut the structure at x=L/2 just before the applied load and keep the left
side.
- I equilibrate acting forces and moments on both sides of the slice:
- Axial force: zero.
- Shear: the acting forces are the same as the ones in the
previous cut, so the result will be the same as well: –P/6.
- Bending moment: the forces now are the same as in the
previous cut, but they are farther this time. If I take moments
about the point at which I’ve cut, the acting forces produce a
clockwise, PL/6 kNm moment, so the bending moment is positive.
- Don’t forget to write down the obtained values on your diagrams.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. Second example (cont).
X = L/2 (just after the point load)
- I split the structure in two at x=L/2 just after the point load and keep the
left side of the cut.
- I equilibrate the forces and moments on both sides of the slice:
- Axial force: zero.
- Shear: The total amount of load perpendicular to the bar is P/6,
downwards. On the right side of the slice I should have a force
going upwards to compensate it, and that’s precisely what the
positive sign criteria is providing me. Therefore the sign is
positive.
- Bending moment: the new force is applied at the point about
which I’m taking moments and the existing forces are at the same
distance… so the value of the moment is the same one as in the
previous cut: +PL/6.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. Second example (cont).
X = 3L/4 (just before the point load)
- I cut the structure at x=3L/4, just before the third point load and keep the left side.
- I equilibrate forces and moments on both sides of the cut:
- Axial force: zero
- Shear: +P/6
- Bending moment: +PL/4
- I write these values down on my diagrams
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. Second example (cont).
X = 3L/4 (just after the third point load)
- I cut the structure at x=3L/4, just after the new point load and keep the left side.
- I equilibrate forces and moments on both sides of the slice:
- Axial force: zero.
- Shear: + P/2.
- Bending moment: +PL/8
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. Second example (cont).
X = L
- I cut the structure at x=L, just before the right support
- I equilibrate forces and moments on both sides of the slice:
- Axial force: zero.
- Shear: +P/2.
- Bending moment: zero.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. Second example (cont).
- To draw the diagrams, I simply join the dots and extract some conclusions.
- Shear values change when new forces perpendicular to the bar appear.
- Bending moment varies linearly: the farther I cut from the force that produces the moment, the bigger the produced
moment is. Whenever new forces appear, the slope of the diagram changes.
- Conclusion: I only need to cut the members near the points where point forces are applied.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. Second example (cont).
- To draw the diagrams, I simply join the dots and extract some conclusions.
- Shear values change when new forces perpendicular to the bar appear.
- Bending moment varies linearly: the farther I cut from the force that produces the moment, the bigger the produced
moment is. Whenever new forces appear, the slope of the diagram changes.
- Conclusion: I only need to cut the members near the points where point forces are applied.
Don’t miss this!
- The “skips” that appear in the shear diagram correspond to the
value of the applied loads at that points.
- Whenever a point load is applied, a skip is produced in the
shear diagram and a change of slope happens in the bending
moment one.
- Positive values in the bending moment diagram represent areas
of the member in which the lower side of the slice is subjected to
tension.
- Bending moment diagrams remember us of the deflection of the
bar.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. Third example
Simply supported beam subjected to uniform, distributed load.
Obtain reactions.
Obtain the values of the internal forces before and after each applied load.
Plot the axial, shear and bending moment diagrams.
.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. Third example
Simply supported beam subjected to uniform, distributed load.
Obtain reactions.
Obtain the values of the internal forces before and after each applied load.
Plot the axial, shear and bending moment diagrams.
.
We already now that we only have to
obtain values at the points where new
loads appear.
In this case, as the load is evenly
distributed, we will obtain an
intermediate value just to deduce
how the variation in the diagrams is
produced.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. Third example (cont).
X = 0
- I cut the structure in two infinitely near the left support and I keep the smallest part
(this would be the last slice before the support). Once I cut, I can see the forces on the
right side of the slice, which will be represented according to the positive sign criteria
for a horizontal bar.
- I equilibrate the remaining forces and moments on the structure according to the
direction of each of my unknowns:
- Axial force: There are no forces in the direction of the axial force for “N” to
equilibrate, so the value will be zero.
- Shear: The only acting force in the shear direction (perpendicular to the axis of the
member) is the vertical reaction. Therefore, the magnitude of the shear at the selected
point of the member is q*L/2. We still have to decide on the sign. If we keep “V” in the
sense it would have if it were positive, it’s easy to see that it wouldn’t be able to
equilibrate the left reaction. This means that the sense of the shear force at that side of
the slice must be opposite to the one we had supposed. Thus, the shear force is
negative.
- Bending moment. We perform sum of moments at the point we have cut. All acting
forces pass through the selected point, so the bending moment at that point has to be
zero.
- Finally, I represent on each diagram the obtained values.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. Third example (cont).
X = L (just before the right support)
- I cut the member at x=L, just before the right support and keep the left side.
- I equilibrate forces and moments on both sides of the slice:
- Axial force: zero.
- Shear: + q*L/2
- Bending moment: zero.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Bending moment: There are two forces producing moments about the point I have cut at: the reaction at the pinned support and the
distributed load until that point (a total load of q*L/2 times the distance: q*L/4). Therefore, the magnitude of the moment produced by
these forces is:
and it’s a clockwise moment
To compensate a clockwise moment on the left of the slice we need a counterclockwise one on the right. This coincidences with my
positive sign criteria. Therefore the bending moment is positive.
Internal forces diagrams. Third example (cont).
Intermediate point: X = L/2
- I cut the member at x=L/2 in order to understand how internal forces
vary between x=0 and x=L.
- I equilibrate forces and moments on both sides of the slice:
- Axial force: N=0
- Shear: the resultant of forces on the left side of the slice in the
direction of the shear is zero. If I think of the variation of the resultant of
the forces, I can easily see that this variation is linear (the more
distance, the more load).
2 1
2 2 2 4 8
q L L L Lq qL
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Bending moment: There are two forces producing moments about the point I have cut at: the reaction at the pinned support and the
distributed load until that point (a total load of q*L/2 times the distance: q*L/4). Therefore, the magnitude of the moment produced by
these forces is:
and it’s a clockwise moment
To compensate a clockwise moment on the left of the slice we need a counterclockwise one on the right. This coincidences with my
positive sign criteria. Therefore the bending moment is positive.
Internal forces diagrams. Third example (cont).
Intermediate point: X = L/2
- I cut the member at x=L/2 in order to understand how internal forces
vary between x=0 and x=L.
- I equilibrate forces and moments on both sides of the slice:
- Axial force: N=0
- Shear: the resultant of forces on the left side of the slice in the
direction of the shear is zero. If I think of the variation of the resultant of
the forces, I can easily see that this variation is linear (the more
distance, the more load).
2 1
2 2 2 4 8
q L L L Lq qL
Don’t miss this!
The variation of the shear is linearly dependent on the
distance, hence the function that represents these values
must be a line.
On the other hand, the variation of the bending moment due
to distributed, uniform load depends on the distance,
squared.
Therefore the function is not linear, but exponential.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. Third example (cont).
Plotting diagrams.
- I join the dots and extract some conclusions.
- Uniform, distributed loads make the shear function to be represented with a line. The slope tells me how much load is
acting per unit of length..
- Uniform, distributed loads make the bending moment function to become a parabolic function. When the shear diagram
reaches zero, I’ll find a change in the sign of the slope of the parabola (therefore, there’s a maximum or a minimum).
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. Third example (cont).
Plotting diagrams.
- I join the dots and extract some conclusions.
- Uniform, distributed loads make the shear function to be represented with a line. The slope tells me how much load is
acting per unit of length..
- Uniform, distributed loads make the bending moment function to become a parabolic function. When the shear diagram
reaches zero, I’ll find a change in the sign of the slope of the parabola (therefore, there’s a maximum or a minimum).
Don’t miss this!
- When a member is subjected to uniform, distributed load, the shear
diagram varies linearly and the bending moment does it exponentially (in
second degree).
- The curvature of the bending moment diagram is similar to the one that a
piece of fabric would have if I blew against it in the direction of the acting
force).
- If an structural system is symmetrical (both, geometry and loads), the
bending moment diagram is symmetrical as well and the shear force
diagram is antisymmetrical (one of the sides is symmetrical and reflected).
- Bending moment diagram reminds me of the deflection of the structure..
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. Third example (cont).
When structural systems become more complex, each of the members will have its own internal
forces diagram. You only must remember that each bar is the x axis and the axis perpendicular to
“x” is the internal force axis.
Check carefully the sign criteria for vertical and inclined members.
Solid Mechanics – Academic Year 2014/2015 Prof: Maribel Castilla Heredia @maribelcastilla v.1.0 October 2014
Internal forces diagrams. Third example (cont).
Sign criteria along this course:
You can see that beside the slice there’s a tiny (+). That
will remind you on which side of the element you should
represent the positive values.
o1. Horizontal member: positive values for internal forces under the bar.
o2. Vertical member: positive values for internal forces on the right side
of the bar.
o3. Inclined member (both positive or negative slope): positive values for
internal forces under the bar.
Example ->.
Solid Mechanics Academic Year 2014/2015
Solid Mechanics
Block A. Internal forces diagrams.
Prof. Maribel Castilla Heredia @maribelcastilla Version 1.0 October 2014
Degree in Architecture.
San Pablo CEU University – Institute of Technology.