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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