Good morning!

New wordsBeam 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

TensionCompression

ShearBendingTorsion

Internal forcesFor 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

CalculationThe SF at any section A in a straight beam is the algebraic sum of all vertical forces lefts of that point

SFA=∑VLA

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 ① ② ③

Bending Moment

BML=∑ALL

•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 ① ② 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 byeSee you next

class