# Good morning!

Urban and Civil

Nov 26, 2013 (4 years and 5 months ago)

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Good morning!

New words

Beam

Tension

Compression

Shear

Torsion

Bending

Concave side

Convex side

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

timber

steel

concrete

New member

beam

Question:

in considering the beam what
are we concerned with?

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…

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 the start and end of every

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

Let’s talk about it on next
lesson!

Homework:

Exersize1

a/b/c/f

Bye bye

See you next class