Good morning!

New words

Beam

梁

Tension

拉伸

Compression

压

Shear

剪

Torsion

扭转

Bending

弯曲

Concave side

凹面

Convex side

凸面

Exersize

•

measure the tendency of one part

of a beam to be slided with respect

to the other part

•

-------

(shear force)

•

the tendency for one part to be

rotated with respect to the other

part

•

-------

(bending moment)

Review:

In last week, we have learned

columns made of :

timber

steel

concrete

New member

–

beam

Question:

in considering the beam what

are we concerned with?

Answer:

the effects of the forces

Five forms of deformation

Tension

Compression

Shear

Bending

Torsion

Internal forces

For example:

the internal forces that

result from bending

deformations are compressive

on the concave side and

tensile on the convex side

•

SF&BM measure the tendency

of one part of a beam to be

slided with respect to the

other part (SF effect),and

the tendency for one part to

be rotated with respect to

the other part(BM effect)

•

The question we need resolve

is to calculate the SF&BM

Shear Force

Calculation

The SF at any section A in a

straight beam is the

algebraic sum of all vertical

forces lefts of that point

SF

A

=∑V

L

A

Eg1:

3KN 5KN

A B C D

2M 2M 1M

8KN

STEPS

•

Cut the beam at point B ,so

we will get part of the

shear force of the beam

8KN 3KN

A B

If we cut the beam at point

C ,what will happen to the SF?

TRYING…

What about point D?

TRYING…

SFD

•

If we plot the value of the

SF at all point along the

axis of a beam, we’ll obtain

the Shear Force Diagram

8KN 5KN

Principle for solving the SFD

•

Calculate the shear forces at the

following significant positions:

–

At the start and end of the beam

–

At every support

–

At every point load

–

At the start and end of every

distributed load

Conclusion:

•

We find that at a point load

the SF changes

instantaneously and the SF

diagram shows a step.

Eg2:

A B C D

2M 2M 1M

Steps:

•

First calculate the reaction at

support A.

•

Then cut the beam at point D

•

Finally plot the SFD

15

A B C D

2M 2M 1M

Conclusion:

•

When there is an UDL,we see

the SF change by equal

intervals ,and the SFD will

have a constant slope.

Have a Try

4KN 6KN 3KN

A

①

†††††††††

②

†††††

B†††††††††

③

Bending Moment

BM

L

=∑A

L

L

•

The BM at any section A in a

straight beam is the algebraic

sum of areas of the SFD left

of that point

Eg3

A B C D

2M 2M 1M

8KN 5KN

A B C D

NOTE

•

Locate the positions of zero

SF, as these will also be

positions of maximum BM.

The result is:

0

16KN

26KN

HAVE A TRY

4KN 6KN 3KN

A

①

†††††††††

②

†††††

B† †††††††

③

CONSIDER

•

All we learned today is the

simply supported beam, then

what about continuous beam?

•

Let’s talk about it on next

lesson!

Homework:

Exersize1

a/b/c/f

Bye bye

See you next class

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