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#### 梁的剪切应力模型

Shear in Beams Model

This model was designed to help students understand

how shear stresses arise in beams,

why those stresses vary from place to place over a beam cross section,

and why they produce specific shear flow patterns.

To provide context for the model,

you might imagine that it is part of a cantilever beam

like this one

If that were the case,

the bending moment acting on the model at section B

would be larger than that acting at section A.

We represent the tensile bending stresses that these moments would produce

using red arrows overlaid on Plexiglas blocks

and the compressive ones

with blue arrows overlaid on the beam.

As you can see

the bending stresses at B are also larger than those at A.
B段的弯曲应力同样比A段的大
As we will discover shortly

differences in bending stresses from one section

to another along the length of a beam

are the reason that shear stresses arise.

Before we begin that discussion, however,

notice that the length of each arrow is proportional

to the magnitude of the stress that it represents.

As a result, the volume of each Plexiglas block

is proportional to the force generated

over its base area.

As you can see,

large forces are generated in the flanges

and smaller ones in the web.

Not only that, but the flange forces are

relatively far from the neutral axis of the beam,

and so they contribute much to the bending moment.

The forces on the web are much smaller

and they act at a shorter moment arm.

As a result, they contribute relatively little to the moment.

To understand how shear stresses arise in a beam,

imagine that we used a saw

As you can see, the force from the bending stresses on end B of the fiber

is larger than that produced at end A.

In the worked example included at the end of this video,

the difference in these forces is 15 units.

So, for this fiber to remain in equilibrium,

the rest of the beam must exert a force of 15 units on it.

We treat the fiber as if it were a free body,

and therefore draw the arrow on the fiber

in the direction of the force that [the beam] exerts on it.

At the same time that the beam is exerting a force of 15 units on the fiber,

the fiber is exerting an equal and opposite force on the rest of the beam,

as shown by this arrow.

As you can see,

the forces between the fiber and the rest of the beam

act parallel to the cut,

and so we call them shear forces,

and we represent them using arrows that have a single-sided head.

In contrast, forces that act normal to a surface,

like those produced by these bending stresses,

and we represent them using arrows that have symmetrical heads.

If we now imagine a cut that removes two fibers of the flange,

that pair will be out of balance in the axial direction by twice

as much as the single fiber we originally considered.

So, to keep this new free body in equilibrium,

a shear force of 30 units must act,

as shown by these arrows.

If we cut off all 5 of the fibers

that make up the top flange from the rest of the beam

the total axial imbalance is 5 times 15

or 75 units.

And that is the magnitude of the shear force

that must act on the newly cut surface

that is between them and the rest of the beam.

A fiber closer to the neutral axis of the beam

experiences bending stresses that are smaller.

The difference between the forces on the two ends of this fiber

is only 9 force units.

Adding 9 to 75 gives a total shear of 84 units.
9加75即得到84单位的总的剪切应力
The difference in the end forces for the fiber just above the

neutral axis is only 3 units.

Adding that to 84 gives a total of 87 units.

Fibers below the neutral axis

carry compression, rather than tension.

As a result, those fibers are out of balance in the opposite axial direction

compared to those above it.

If we imagine a cut below the neutral axis

some of these fibers are now included

and they reduce the total axial imbalance

compared to a beam cut right at the neutral axis.

That is why the maximum shear force

is always found at the neutral axis.

Notice that these shear forces arise

because the bending stresses

and the moments that cause them

vary with position along the length of the beam.

It is customary to report the shear stresses on a beam cross-section

rather than the shear forces.

For the sake of simplicity,

the web and flanges in this model are assumed to be of unit thickness.

Also, the two cross-sections are considered to be separated by a unit amount,

even though the physical dimensions of the model may suggest otherwise.

As a result of these dimensional choices,

the shear stress along any of the cuts we have made

will be numerically equal to the shear force it carries.

Notice too that our cuts always

go across the thickness of the web or flanges.

That way all of the points in the cut are very near to each other

and would generally be expected to carry similar stresses.

We never cut the flange parallel to its top surface, for example.

Points along such a cut would be relatively distant from each other

and the stresses at various points along the cut could differ substantially.

One can plot the magnitude of the shear stress

as a function of the distance between the cut and the outer extremity of the flange.

As you can see, the graph is linear,

a result confirmed by the calculations

shown on the right.

for other regions of the flanges

and, by convention,

the axes are typically oriented as shown here.

As these calculations show,

the shear stress in the web takes a parabolic form.

Plotting all of these graphs on a single figure reveals what is called

the “shear stress distribution” for that cross section.

In order to understand shear flow,

we must transfer the shear stresses we just calculated

from their respective longitudinal cutting planes, to the beam cross section.

We do that using the simple fact that

shear in a plane always involves four matched stresses

If we concentrate on this inside corner of a flange fiber,

the shear on the cross-section must go tail-to-tail

with the longitudinal stress arrow

and so it must point outwards, and its magnitude must be 15.

If we focused on the outside corner of the adjacent flange fiber,

we would get exactly the same result.

We can show the shear stress on the cross section using a single-sided arrow.

At two fibers in, the shear stress is 30,

and so that is the stress shown on the cross section at that location.

To analyze the web, we note that its shear operates in a vertical plane,

unlike that in the flanges, which operates in a horizontal plane.

Collectively, these shear arrows

show how the shear flows over the cross section,

and when taken together,

they reveal what is known as the “shear flow”.

These arrows represent forces or stresses that must act on the beam

for it to remain in equilibrium.

So, clearly, an upwards external force must act

on the end of the beam that carries the larger bending stresses

And a downwards force must act on the other end.

If you examine the beam from which this model was taken,

you can see that those shear directions are indeed correct.

Not only that,

but if we estimate the total vertical load on the beam cross section

by summing the indicated vertical forces

we get 405 units.

This value is quite close to the shear of 412 units

noted in the worked example that follows the credits.
412是实际工作中所记录的权威数字
It may seem surprising that the shear forces acting on a beam

could be determined solely from the bending stresses

acting on two nearby cross sections.

However, the differences between the stresses on those two sections

are a direct result of changes in the bending moment M

with axial position x.

And that rate of change can be used to calculate the beam shear

by using the well-known formula, V=dM/dx.
V＝dM/dx来计算剪切应力
This relationship explains, mathematically,

why shear and changes in bending moment are so closely related.

We hope this video helped you to understand

how shear stresses arise in beams,

why they vary over a beam cross section,

and how they produce specific shear flow patterns.

Thanks for watching.