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







Shear force




parallel.







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







Shear force




parallel.







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.













Axial force







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.













Axial force







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:










- 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.


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










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










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.



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.




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

- 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.





- 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.




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






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 (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





- 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.

Obtain reactions.



.


Internal forces diagrams. Third example

Simply supported beam subjected to uniform, distributed load.

Obtain reactions.



.

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

- 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.





- 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.




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.


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 Lq qL


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 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.



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 ->.

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.

Bending moment, shear and normal diagrams

Education

Transcript of Bending moment, shear and normal diagrams