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plantcitybusinessUrban and Civil

Nov 26, 2013 (3 years and 9 months ago)

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Good morning!

New words

Beam


Tension
拉伸

Compression


Shear


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

†††††††††

†††††
B†††††††††




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

†††††††††

†††††
B† †††††††




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