Bending moment, shear and normal diagrams

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:

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

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.

Internal forces
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.

Internal forces
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.

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.

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.

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.

Horizontal member

Horizontal member
Vertical member

Horizontal member
Inclined member. Increasing slope.
Vertical member

Horizontal member
Inclined member. Increasing slope.
Vertical member
Inclined member. Decreasing slope.

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.

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.

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.

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.

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.

X = 0
- 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.

X = 0
- 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.

X = 0
- 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.

X = 0
- I equilibrate the remaining forces on the structure according to the direction of each of my unknowns:
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.

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:

X = L/4
- 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 .

X = L/4
- 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.

X = L/4
- I represent the obtained values on the diagrams.

X = L/2 (before the point load)
- I split the bar in two at x=L/2 and I keep the left part.
- 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.

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:
- 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.

- I equilibrate forces and moments on one side and on the other side of the slice:
- 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.

- I equilibrate the forces and moments one side and on the other side of the slice:
- 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.

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:
- 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

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.

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.

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.

Internal forces diagrams. Second example (cont).
X = 0

X = L/4 (just before the first applied load)
- 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:

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:
- 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.

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:
- 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.

- 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:
- 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.

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

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:
- Shear: + P/2.
- Bending moment: +PL/8

X = L
- I cut the structure at x=L, just before the right support
- Shear: +P/2.
- Bending moment: zero.

- To draw the diagrams, I simply join the dots and extract some conclusions.

- To draw the diagrams, I simply join the dots and extract some conclusions.
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.

Internal forces diagrams. Third example
Simply supported beam subjected to uniform, distributed load.
Plot the axial, shear and bending moment diagrams. .

Internal forces diagrams. Third example
Simply supported beam subjected to uniform, distributed 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.

Internal forces diagrams. Third example (cont).
X = 0
- 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.

X = L (just before the right support)
- I cut the member at x=L, just before the right support and keep the left side.
- 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.
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.
- 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 L
q 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.
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.
- 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 L
q 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.

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).

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..

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.

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 ->.

Bending moment, shear and normal diagrams

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Viewers also liked

Viewers also liked (20)

Similar to Bending moment, shear and normal diagrams

Similar to Bending moment, shear and normal diagrams (20)

Recently uploaded

Recently uploaded (20)

Bending moment, shear and normal diagrams