Shear Force and Bending Moments: Shear force and Bending Moment Diagrams

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Do curso por Georgia Institute of Technology

Fundamentos da Engenharia - Revisão

176 classificações

Georgia Institute of Technology

Fundamentos da Engenharia - Revisão

176 classificações

The purpose of this course is to review the material covered in the Fundamentals of Engineering (FE) exam to enable the student to pass it. It will be presented in modules corresponding to the FE topics, particularly those in Civil and Mechanical Engineering. Each module will review main concepts, illustrate them with examples, and provide extensive practice problems.

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Mechanics of Materials

This module reviews the principles of the mechanics of deformable bodies. We first review the basic concepts of equilibrium and stresses and strains in prismatic bars under axial loading. Then we discuss the major mechanical properties of common engineering materials, particularly the diagrams for normal stress and strain leading to Hooke’s Law, and their relation to lateral strain through Poisson’s ratio. Shear stresses and their relation to shear strains are then presented. We then analyze in detail deformations and stresses in axially loaded members. This includes uniform and nonuniform loading for statically determinate and indeterminate structures. Thermal effects are then considered: expansion and contraction under temperature changes and the stresses that may develop both with and without prestresses. Stresses on inclined planes under axial loadings and the resulting maximum and minimum normal and shear stresses that result are then discussed. Torsion, the twisting of circular rods and shafts by applied torques is then analyzed. We show how to calculate the angle of twist and shear stress as functions of rod properties and shape under uniform and nonuniform torsion. Applications to power transmission by rotating shafts are presented. We then discuss how shear forces and bending moments arise in beams subject to various loading types and how to calculate them. This is then generalized to local forms of the equilibrium equations leading to rules for drawing shear force and bending moment diagrams. Finally, we compute bending stresses in beams. Strains due to bending and their relation to curvature are first discussed. This is used to compute the bending stresses and their relation to the applied bending moment and beam material and cross sectional properties. This includes a review of computation of centroids and moments of inertia of various areal shapes. We complete this module with a discussion how shear stresses arise in beams subject to nonuniform bending and how to compute them. In all cases, basic ideas and equations are presented along with sample problems that illustrate the major ideas and provide practice on expected exam questions. Time: Approximately 4 hours | Difficulty Level: Medium

Stresses and Strains: Introduction 14:23

Stresses and Strains: Mechanical Properties 13:17

Stresses and Strains: Shear Stress 13:08

Axial Loadings: Axial Loaded Members 14:01

Axial Loadings: Statically Indeterminate Structures10:04

Axial Loadings: Thermal Effects and Stresses on Inclined Surfaces 18:45

Torsion: Circular Bars in Pure Torsion19:57

Torsion: Nonuniform Torsion and Power 16:25

Shear Force and Bending Moments: Introduction to Bending of Beams17:11

Shear Force and Bending Moments: Shear force and Bending Moment Diagrams 18:39

Stresses in Beams: Strains in Pure and Nonuniform Bending12:35

Stresses in Beams: Strains in Pure and Nonuniform Bending (continued)4:47

Stresses in Beams: Stresses, Moment-Curvature Equation, and Geometric Properties10:07

Stresses in Beams: Digression (Centroids and Moments of Areas)4:41

Conheça os instrutores

Dr. Philip Roberts
Professor
School of Civil and Environmental Engineering

I want to continue our discussion of shear forces and bending moments.

And in this segment, we'll look at the local form of the equilibrium equations.

And show how to use them to draw shear force, and

bending moment diagrams in beams.

Now, firstly before we do that, let me remind you of the sine convention

that we assumed for sheer forces in bending moments, look like this.

So, these are positive on the left-hand side, and negative on the left-hand side.

Or, separating them out, the sign convention for

positive sheer forces looks like that.

In other words they're trying to rotate the element around

in a clockwise direction.

And the positive sign convention for bending moment looks like that.

In other words, the beam is being compressed at the top.

And elongated, or in tension at the bottom.

Now let's first illustrate this by means of this example.

I have a simple beam which is supporting two loads as shown her, and a couple.

And the question is, find the shear force and the bending moment.

Just very slightly to the left of the midpoint.

In other words, to the left of the point where that applied moment is applied, and

just very slightly to the right of it.

So first of all, we apply statics to the whole beam to find the reactions

RA RB and that's a simple straightforward statics problem.

And the result you should get is this.

RA is 3P over 4 minus M0 over L, and RB is equal to that.

Now, first I'm going to make a cut here, just to the left of the midpoint.

In other words, just to the left of where that applied couple is applied.

Okay so the free body diagram looks like that.

Where V is the shear force and M is the bending moment across that cut.

Where their directions are assumed to be in the positive direction,

according to our sign convention.

So, first of all,

if I sum the forces in the vertical direction to get the sheer force.

The forces in the vertical we have are RA, P, and V.

In other words, RA minus P minus V is equal to zero.

Or the shear force is equal to minus P over 4, minus M0 over L.

Next, I take moments about the cut.

So I'm going to take moments about here, and

I'll assume that counterclockwise direction is positive.

And again, it doesn't matter which direction you assume here.

Summing the cuts about that point equal to zero, for equilibrium.

We have a known moment here, M.

We have the moment of P, and the moment of Ra.

And solving that, we find that the moment there, the bending moment,

is equal to PL over 8, minus M zero, over 2.

Next I'll make a cut just to the right of the mid point.

So, now I'm going to make my cut right here.

Just to the right of that applied moment.

And here is my new free body diagram.

Which is similar to the first one.

Except now we have this additional concentrated moment.

M0 applied.

So I do the same analysis.

First of all, sum the forces in the vertical direction.

And that leads to exactly the same equation, and

the same result for the shear force V.

Now I take moments about this point.

Summation of the moments about the cart is equal to zero.

And filling in the data the details there.

We find that M, the bending moment, is equal to PL over 8 plus M0 over 2.

So, there's a couple things we notice about that.

That these two sheer forces here, just to the left and

the right of that cut, are the same.

The is no sudden discontinuity in share force across that concentrated moment.

However, the two results for the bending moment are different.

So there is a sudden change in the bending moment in the beam,

as we pass over that concentrated moment.

And it will turn out that the change in the bending moment across that point,

is equal to the magnitude of the applied concentrated moment.

Now, what I want to do is to generalize these results.

And see how these things vary as we change.

As we go across a concentrated force.

How the various distributed loads, etc.

So, what we want to do is to derive local forms of equilibrium relationships,

for different types of loading as shown here.

For example a distributed load.

A concentrated load, or a concentrated couple.

And here I will only give the results.

First of all, for distributed loads, we do an analyses like that.

We have a distributed load of intensity q,

over a small element of the beam of horizontal length dx.

And I'll suppose at the moment on the left hand side is M.

And the moment on the right hand side is slightly different, is M plus DM.

Similarly, the sheer force on the left hand side is V.

But the sheer force on the right hand side is different by a small amount, dV.

By applying equilibrium to this element, I find these results that dV by the dx.

The rate of the change of the share force along the beam,

is equal to the negative of the applied loaded intensity.

And the rate of change of the bending moment dM by dx,

is equal to the local shear force V.

Next, for concentrated loads.

And now our element looks like this.

I have a concentrated load P at the top.

And now I'm going to assume that

the bending moment change was by a finite amount m1.

And the sheer force changes by a finite amount V1.

So, similar analysis summing the forces in the moments on this element.

We find that V1,

the change in the sheer force is equal to the negative of the applied load.

Where we're assuming that the applied load is downwards.

And the change in the bending moment across there is equal to 0.

M1 is equal to 0.

In other words M is constant.

It doesn't change across the concentrated load.

Which is what we found in the previous example.

dm by dx does however change.

The slope of the bending moment does change across the finite consecrated load.

And the last one is a concentrated moment, or couple.

Which looks like this.

Now we have a concentrated moment, M0, applied here in this direction.

And I'll assume that the change in the moment across this element is M1.

And the change in the sheer force is V1.

Applying equilibrium to that element.

We find that V1 is equal to 0.

In other words, the sheer force is constant.

It's not changed as a result of this applied moment.

But m1 is equal to -m0.

In other words, the bending moment in the beam changes by a finite amount.

It jumps by an amount which is negative of the applied moment.

According to the sign conventions here.

So, these relationships give rise to rules

that allow us to draw diagrams of shear force and bending moments.

Which will in turn help us to find and see where the maximum occur in a beam.

So let's see how to do that.

So here's a generalization of those equations that I just gave.

So the first rule, is that the slope of the shear diagram, is the negative

of the slope of the shear force intensity, dV/dx=-q(x).

It follows from that that if I integrate this between two points A and B.

That the change in shear force between any two points.

A and B.

Is the negative of the area under the shear force diagram.

Next, the slope of the bending moment equals the local shear force.

dM by dx is equal to V.

Number four, the change in the bending moment

between two points equals the area under the shear diagram.

Which follows simply by integrating this expression between two points.

5.

The moment is a local maximum or minimum.

Where the shear force changes sign.

In other words from positive to negative or negative to positive.

6.

Concentrated forces cause a jump in the shear force

which is equal to the magnitude of the shear force.

V1 = -P.

And concentrated moments cause a jump in the moment equal

to the magnitude of the moment, M1= -M0.

And the minus signs here of course just follow from the sign conventions that

we've adopted.

And here is the relevant section from the reference manual.

So, let's do a couple of examples, and

this is summary of some of those equations.

And let's first of all look at a concentrated load.

So I have a simply supported beam.

A pen support at the left hand side, and a roller support at the right hand side.

With a concentrated load P, at this point.

How do those rules apply?

If I draw the sheer diagram, it looks like that.

So, it's easiest to imagine that I start from just to the left of the point A here.

The sheer force to the left of there is zero.

Then I have an upward reaction force here, RA.

So I can think of that as a concentrated load.

And that concentrated load, causes a jump in the sheer force as I go across there.

So, the jump according to this equation,

is actually in the same direction as the arrow.

So, this concentrated force causes the sheer force to jump up.

By an amount equal to the reaction RA which is actually equal PB/L.

Then, as I move to the right, we see that dV by dx is equal to minus q of x.

However, there is no distributed load here.

So, q of x is zero.

In other words, dV by dx is zero.

In other words, V is constant.

So the shear force is constant in this region.

Then we get to the next point here, the concentrated load P.

And this causes the shear force to drop down by a distance equal to that force P.

So we drop down here by an amount P.

So this height here is P to this point.

And then along here, again, the distributed load is zero.

So the sheer force is constant.

And then finally we get to the reaction R B.

Which brings us back to the beginning to here.

So that is the sheer force style.

Then the bending moment diagram, looks like this.

So here we can see that we have dM by dx.

The rate of change of the shear force of the bending moment,

is equal to the local shear force.

And the shear force here is constant Therefore dm by dx is constant.

In other words, M increases linearly.

So M is zero at the left-hand side here, and

increases linearly up to this point here.

As we go across this point,

the concentrated force there, or concentrated load.

Doesn't cause any change in the bending moment.

So there isn't a discontinuity in the bending moment at that point.

But now the sheer force is negative, and the bending moment decreases back to zero.

So, remembering here we can also think of that In terms of this equation.

That the change in the bending moment between two points,

is equal to the area under the shear force diagram.

In other words, the bending moment at this point is equal to this area right here.

And, in addition, here the area is negative.

In other words the bending moment is zero.

So this area must be equal to this area.

So again we see that there is not a discontinuity in the bending moment at

this point.

However, there is a discontinuity in the slope of the bending moment at that point.

as a result of the concentrated load.

The second example on the right here has a uniformly distributed load.

And firstly, from statics we can solve for

the reactions, and also the sheer force.

So now the sheer force at any location here Is varying across there.

And it's given by this equation.

Where X is the distance measured from the left hand side.

And similarly from statics, the bending moment is given by this expression.

Which it can fairly easily derive.

So, if I plot those out, they look like this.

Now, in this case, we have an upward concentrated load here.

So the sheer force jumps up by that amount.

Which is equal to qL over two.

But now we see dV by dx.

Here, dV by dx is minus q of x.

And q is pushing downwards.

So dV by dx is negative.

So therefore the shear force is dropping as we showed from the equation.

This continues to drop until the right hand side.

When it meets the upward concentrated load.

Which brings the shear force back to zero.

And similarly, if we plot out the bending moment diagram.

We get a relationship which looks like this, which is now quadratic.

It's bending like this.

So again, the area

under the shear force diagram here, is equal to the bending moment.

So the bending moment here is equal to this shaded area.

And from this point onwards, the area is negative.

Because the shear force is negative.

Therefore, the bending moment is decreasing back to zero.

So again, these two areas must be the same.

And the other thing we notice,

is that the maximum in the bending moment here occurs at this point.

It occurs where the sheer force changes sign from positive to negative.

And another one concentrated loads.

You can study that one, it must look like that.

And the sheer force diagram looks like that.

So, again the bending moments here are continuous.

They don't suddenly change at a concentrated load.

However.

The slope changes suddenly, at this point.

And again, you see that the maximum in the bending moment here occurs, at this point.

Where the sheer force changes sign from positive to negative.

Or it could be at a location from negative to positive.

But where it changes sign.

And another one a little bit more complicated example.

But still a simply supported beam subject to a distributed load and

concentrated moment.

And the shear force diagram looks like that.

So, the sheer force is changing here under the distributed load.

But it's constant over this region here where there is no distributed load.

And the occurrence of the concentrated couple doesn't change it.

The bending moment diagram looks like this.

And here, see you have to be a little bit careful.

Because in this case, the maximum bending moment

doesn't occur at the location of an applied concentrated force.

In this case, the maximum bending moment

occurs at the location of the applied concentration couple.

But that may not necessarily be the case.

We have to be careful that the maximum bending moment

doesn't always occur at the location of a concentrated load.

So, this shows us how to use these

elementary rules to draw sheer force and bending moments.

Which will help us to spot, and see where the maximum sheer force, or

the maximum bending moments occur in a beam.

Which is what we're generally interested in in terms of beam design

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